Q.Find dxdy in the following: x=acosθ,y=bcosθ
Concept understanding — Implicit Differentiation
Implicit Differentiation
When y isn't alone
You can differentiate y=x2+3x term by term because y is written explicitly in terms of x. But an equation like x2+y2=25, or x3+y3=6xy, does not give y by itself — solving for y is messy or downright impossible.
Implicit differentiation finds dxdy without isolating y: treat y as an unknown function of x, differentiate the whole equation as it stands, then solve for dxdy.
The one key move: y is really y(x)
Wherever y appears, picture y(x) hiding inside. Differentiating a y-term therefore needs the chain rule, which tacks on a factor of dxdy:
dxd(y2)=2ydxdy.
That extra dxdy on every y-term is the whole trick.
The procedure
- Differentiate both sides with respect to x, treating y as y(x).
- Each time you differentiate a y-term, multiply by dxdy (chain rule); use the product rule on mixed terms such as xy.
- Gather all dxdy terms on one side, everything else on the other.
- Factor out dxdy and divide.
Worked example
For x2+y2=25:
2x+2ydxdy=0⇒dxdy=−yx.
The answer naturally contains both x and y — that is normal here. To get the slope at a point on the curve, substitute the coordinates after differentiating; there is no need to solve for y first.
The classic mistake is dropping the dxdy factor — writing dxd(y2)=2y treats y as if it were x. If a term contains y and you are differentiating with respect to x, the chain rule always applies.
Implicit differentiation is not a new rule; it is the chain rule used systematically whenever y is tangled up with x.
Implicit differentiation is a named subtopic of the NCERT Class 12 Continuity and Differentiability chapter and shows up regularly in CBSE board 'find dy/dx' questions involving equations like x² + y² = 25 that can't easily be solved for y. Students searching 'implicit differentiation class 12 examples' or preparing this technique for JEE Main will recognize this as simply the chain rule applied systematically to every y-term.
The key idea is Implicit Differentiation (or parametric differentiation, since both x and y are given in terms of θ).
Step 1: Differentiate each parametric equation with respect to θ:
dθdx=−asinθ,dθdy=−bsinθ
Step 2: Use the chain rule for parametric curves:
dxdy=dx/dθdy/dθ=−asinθ−bsinθ
Step 3: Cancel −sinθ (provided sinθ=0):
dxdy=ab
The derivative is ab.
Both x and y are expressed in terms of θ, so we use parametric differentiation: dxdy=dx/dθdy/dθ. Here, x=acosθ and y=bcosθ, giving dxdy=ab.
When you see two equations like x=acosθ and y=bcosθ, the natural instinct might be to eliminate θ first. But that’s unnecessary here — and would actually obscure the simplicity. Both x and y are already functions of the same parameter θ, so we can differentiate each with respect to θ and then take their ratio. This is the essence of parametric differentiation.
The key idea: if x and y are both given in terms of a third variable (the parameter), then
dxdy=dx/dθdy/dθ,
provided dx/dθ=0. This works because the chain rule lets us cancel dθ like a fraction — but only when both derivatives exist and the denominator is non-zero.
Let’s apply it step by step.
- Differentiate x with respect to θ. x=acosθ The derivative of cosθ is −sinθ, so
dθdx=−asinθ.
- Differentiate y with respect to θ. y=bcosθ Similarly,
dθdy=−bsinθ.
- Take the ratio.
dxdy=dx/dθdy/dθ=−asinθ−bsinθ.
- Simplify. The −sinθ cancels (provided sinθ=0), leaving
dxdy=ab.
A common mistake is to forget the minus signs or to cancel them incorrectly. Here both numerator and denominator have a factor of −sinθ, so they cancel cleanly. But if sinθ=0, the derivative is undefined (the curve has a vertical tangent or a cusp at those points — check θ=0,π,…).
Notice that x and y are both proportional to cosθ. So y=abx — the curve is actually a straight line through the origin! That’s why the derivative is constant: the slope is always b/a, independent of θ.
The derivative is ab.
Method: Differentiating a Parametric Curve
This method applies whenever a curve is given through a third variable (a parameter — commonly t or θ) instead of y written directly as a function of x.
Steps
Step 1: Recognise the parametric form
If you are given x=f(param) and y=g(param) instead of y=h(x), do not try to eliminate the parameter first — it is often messy or impossible. Differentiate each equation separately with respect to the parameter instead.
Step 2: Differentiate x and y with respect to the parameter
Use the ordinary rules (product rule, chain rule, standard derivatives) to find dθdx (or dtdx) and dθdy (or dtdy).
Step 3: Divide — the parametric-derivative formula
dxdy=dx/dθdy/dθ,dθdx=0.
This is justified by the chain rule: dθdy=dxdy⋅dθdx, so dividing recovers dxdy.
Step 4: Simplify with trigonometric identities where possible
Parametric answers built from sin,cos of the parameter very often simplify with a double-angle or half-angle identity (sin2θ=2sinθcosθ, 1−cosθ=2sin22θ, 1+cosθ=2cos22θ, etc.) — always look for one before leaving the answer as a raw ratio.
Applying to this problem: with x=acosθ, y=bcosθ, both derivatives with respect to θ pick up the same factor −sinθ, so it cancels in the ratio, leaving the constant ab — a signal that the curve is really the straight line y=abx, whose slope never depends on θ.
Common Mistakes
Mistake 1: Trying to eliminate θ before differentiating.
Why it's wrong: the elimination is unnecessary extra work here — both x and y already reduce to the same trig function of θ, so the parametric-ratio method gives the slope in one line. Correct approach: differentiate x and y with respect to θ directly and divide.
Mistake 2: Cancelling −sinθ without noting it must be nonzero.
Why it's wrong: at θ=0,π the parametric derivatives are both 0, so the ratio ab formally still holds only away from those points where dx/dθ=0. Correct approach: state the cancellation is valid for sinθ=0.
Showing the 12 most recent of 17 on this concept.
- TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQQ.If x=sin2θcos3θ, y=sin3θcos2θ, then dxdy= (A) 2cos5θ−cos3θcos2θ2cos5θ+sin3θsin2θ (B) 2cos5θ+cos3θcos2θ2cos5θ−sin3θsin2θ (C) 2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ (D) 2cos5θ−cos3θcos2θ2cos5θ−sin3θsin2θ
›Reveal solutionSolution
dxdy=2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ.
Differentiate each with respect to θ:
dθdx=2cos2θcos3θ−3sin2θsin3θ,
dθdy=3cos3θcos2θ−2sin3θsin2θ.
Using cos5θ=cos2θcos3θ−sin2θsin3θ, write each derivative around cos5θ:
dθdx=2(cos2θcos3θ−sin2θsin3θ)−sin3θsin2θ=2cos5θ−sin3θsin2θ,
dθdy=2(cos2θcos3θ−sin2θsin3θ)+cos3θcos2θ=2cos5θ+cos3θcos2θ.
Hence
dxdy=dx/dθdy/dθ=2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ.
✓Final answerANSWER: C — 2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ.
- TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQQ.If x=sin2θcos3θ, y=sin3θcos2θ, then dxdy= (A) 2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ (B) 2cos5θ+cos3θcos2θ2cos5θ+sin3θsin2θ (C) 2cos5θ+cos3θcos2θ2cos5θ−sin3θsin2θ (D) 2cos5θ−sin3θsin2θ2cos5θ−cos3θcos2θ
›Reveal solutionSolution
Parametric differentiation gives dxdy=2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ. Option (A).
Solution
With x=sin2θcos3θ and y=sin3θcos2θ, differentiate each by the product rule:
dθdx=2cos2θcos3θ−3sin2θsin3θ,dθdy=3cos3θcos2θ−2sin3θsin2θ.
Use cos5θ=cos(3θ+2θ)=cos3θcos2θ−sin3θsin2θ, i.e.
2cos5θ=2cos3θcos2θ−2sin3θsin2θ.
Numerator:
dθdy=(2cos3θcos2θ−2sin3θsin2θ)+cos3θcos2θ=2cos5θ+cos3θcos2θ.
Denominator:
dθdx=(2cos3θcos2θ−2sin3θsin2θ)−sin3θsin2θ=2cos5θ−sin3θsin2θ.
Hence
dxdy=dx/dθdy/dθ=2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ.
✓Final answerOption (A): 2cos5θ−sin3θsin2θ2cos5θ+cos3θcos2θ.
- TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQQ.If x=32cos3θ and y=4tan2θ then (dxdy)θ=π/4= (A) 9322 (B) 916 (C) −916 (D) −932
›Reveal solutionSolution
With x=32cos3θ, y=4tan2θ, the parametric derivative reduces to dxdy=92cos5θ−8; at θ=4π this is −932, option (D).
- Differentiate each parameter.
dθdx=32⋅3cos2θ⋅(−sinθ)=−92cos2θsinθ,
dθdy=4⋅2tanθsec2θ=8tanθsec2θ.
- Form the ratio.
dxdy=−92cos2θsinθ8tanθsec2θ.
Since tanθsec2θ=cos3θsinθ,
dxdy=−92cos2θsinθ8cos3θsinθ=92cos5θ−8.
- Evaluate at θ=4π. Here cos4π=21, so cos54π=421. The denominator becomes 92⋅421=49, hence
dxdy=92⋅421−8=9/4−8=−932.
✓Final answerdxdyθ=π/4=−932 — option (D).
ANSWER: D
- TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQQ.If x=cos3θ−sin3θ and y=3cosθ−3sinθ, then the value of dxdy at θ=4π is (A) 9232 (B) 332 (C) 9432 (D) 932
›Reveal solutionSolution
Differentiate x and y separately with respect to the parameter θ and divide. At θ=π/4 this gives dxdy=94⋅2−2/3=9232 — option (A).
The concept first
When both coordinates are given through a parameter, x=x(θ) and y=y(θ), the chain rule gives
dxdy=dx/dθdy/dθ(provided dθdx=0).
Never try to eliminate θ here — with a cube and a cube root in the same problem that would be brutal. Just differentiate each expression in θ and take the quotient at the required value.
The symmetry sin4π=cos4π=21 makes the arithmetic collapse very neatly, so keep the powers of 2 in index form until the end.
Step-by-step
- Differentiate x=cos3θ−sin3θ:
dθdx=3cos2θ(−sinθ)−3sin2θ(cosθ)=−3sinθcosθ(cosθ+sinθ).
- Differentiate y=cos1/3θ−sin1/3θ:
dθdy=31cos−2/3θ(−sinθ)−31sin−2/3θ(cosθ)=−31(sinθcos−2/3θ+cosθsin−2/3θ).
- Put θ=4π, where sinθ=cosθ=2−1/2:
dθdx=−3(21)(22)=−3⋅21⋅2=−232.
dθdy=−31(2⋅2−1/2⋅21/3)=−322−1/2+1/3=−322−1/6.
(Here cos−2/3θ=(2−1/2)−2/3=21/3, and the two terms are identical, hence the factor 2.)
- Divide:
dxdy=−2321/2−322−1/6=32⋅32⋅2−1/6−1/2=942−2/3.
- Tidy the surd. Since 2−2/3=221/3,
dxdy=94⋅221/3=9221/3=9232≈0.28.
✓Final answerdxdyθ=π/4=9232.
ANSWER: A
- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.If y=acos3x+be−x, then y′′(3sin3x−cos3x)= (A) 10y′sin3x+3y(sin3x+3cos3x) (B) 10y′cos3x+3y(sin3x+3cos3x) (C) 10y′cos3x+3y(cos3x+3sin3x) (D) 10y′cos3x+3y(sin3x−3cos3x)
›Reveal solutionSolution
The key idea is to compute the first and second derivatives of y=acos3x+be−x, then substitute into the expression y′′(3sin3x−cos3x) and simplify to match one of the given forms. The result simplifies to 10y′cos3x+3y(sin3x+3cos3x), which corresponds to option (B).
We start with the given function:
y=acos3x+be−x
We need to find an expression for y′′(3sin3x−cos3x) in terms of y and y′. The trick is to avoid solving for a and b explicitly — instead, we differentiate and then cleverly combine terms.
1. Compute the first derivative y′
Differentiate term by term:
y′=−3asin3x−be−x
(Recall: derivative of cos3x is −3sin3x, and derivative of e−x is −e−x.)
2. Compute the second derivative y′′
Differentiate y′:
y′′=−9acos3x+be−x
(Derivative of −3asin3x is −9acos3x; derivative of −be−x is +be−x because −e−x differentiates to e−x.)
3. Form the expression y′′(3sin3x−cos3x)
Substitute y′′:
y′′(3sin3x−cos3x)=(−9acos3x+be−x)(3sin3x−cos3x)
Expand:
=−27acos3xsin3x+9acos23x+3be−xsin3x−be−xcos3x
4. Express everything in terms of y and y′
We have:
y=acos3x+be−x
y′=−3asin3x−be−x
We want to rewrite the expanded expression. Notice the terms:
- 9acos23x can be linked to y and y′ if we also have sin2 terms, but we don't. Instead, we aim to match the pattern in the options, which involve y′cos3x and y(sin3x+3cos3x).
5. Try to express acos3x and be−x from y and y′
From y and y′ we can solve:
y=acos3x+be−x
y′=−3asin3x−be−x
Add them:
y+y′=acos3x−3asin3x=a(cos3x−3sin3x)
So:
a=cos3x−3sin3xy+y′
Also subtract:
y−y′=acos3x+3asin3x+2be−x
But this gets messy. Instead, a better approach: directly compute the target expression using derivatives and compare to options.
6. Compute y′′ in terms of y and y′
Notice that from y′=−3asin3x−be−x, we can differentiate again but also note:
y′′=−9acos3x+be−x
We can write y′′ as a combination of y and y′:
y=acos3x+be−x
y′=−3asin3x−be−x
Multiply y by something and y′ by something to get y′′. Try:
y′′=−9acos3x+be−x=αy+βy′
Substitute:
α(acos3x+be−x)+β(−3asin3x−be−x)=a(αcos3x−3βsin3x)+be−x(α−β)
Compare with y′′=−9acos3x+be−x:
- Coefficient of acos3x: α=−9
- Coefficient of asin3x: −3β=0⇒β=0
- Coefficient of be−x: α−β=−9=1? That gives −9=1, impossible.
So y′′ is not a linear combination of y and y′ alone (because the sine and cosine terms have different dependencies). So we must work directly.
7. Instead, compute y′′(3sin3x−cos3x) by substituting y′′ and then replace acos3x and be−x using y and y′
We have:
y′′(3sin3x−cos3x)=(−9acos3x+be−x)(3sin3x−cos3x)
Expand:
=−27acos3xsin3x+9acos23x+3be−xsin3x−be−xcos3x
Now, note that y′=−3asin3x−be−x, so 3asin3x=−y′−be−x. But we have −27acos3xsin3x=−9cos3x(3asin3x)=−9cos3x(−y′−be−x)=9y′cos3x+9be−xcos3x.
So the expression becomes:
=9y′cos3x+9be−xcos3x+9acos23x+3be−xsin3x−be−xcos3x
Combine the be−xcos3x terms: 9be−xcos3x−be−xcos3x=8be−xcos3x.
So:
=9y′cos3x+9acos23x+3be−xsin3x+8be−xcos3x
8. Now rewrite 9acos23x
We know y=acos3x+be−x, so acos3x=y−be−x. Then:
9acos23x=9cos3x(acos3x)=9cos3x(y−be−x)=9ycos3x−9be−xcos3x
Substitute back:
=9y′cos3x+(9ycos3x−9be−xcos3x)+3be−xsin3x+8be−xcos3x
Combine the be−xcos3x: −9+8=−1, so we get −be−xcos3x.
Thus:
=9y′cos3x+9ycos3x+3be−xsin3x−be−xcos3x
9. Factor be−x terms
Notice 3be−xsin3x−be−xcos3x=be−x(3sin3x−cos3x). But be−x=y−acos3x. However, we can also write be−x=y−acos3x, but we want everything in y and y′. Alternatively, note that y′=−3asin3x−be−x doesn't directly give be−x alone.
Instead, observe that 3ysin3x=3acos3xsin3x+3be−xsin3x. And 9ycos3x we already have. Let's try to match the options: they all have 10y′cos3x+3y(something).
10. Compare with option (B): 10y′cos3x+3y(sin3x+3cos3x)
Expand option (B):
10y′cos3x+3ysin3x+9ycos3x
Our expression is:
9y′cos3x+9ycos3x+3be−xsin3x−be−xcos3x
We need to turn 9y′cos3x into 10y′cos3x and the remaining terms into 3ysin3x. That suggests y′cos3x appears with coefficient 10, so we might have missed a y′cos3x term. Let's re-check step 7: we had 9y′cos3x from the manipulation, but perhaps there's an extra y′cos3x hidden.
11. Alternative direct substitution
Let’s compute y′′(3sin3x−cos3x) by writing y′′ in terms of y and y′ using the original differential equation. Since y=acos3x+be−x, note that y satisfies a linear ODE. Differentiate twice: y′=−3asin3x−be−x, y′′=−9acos3x+be−x. Add 9y:
y′′+9y=(−9acos3x+be−x)+9(acos3x+be−x)=10be−x
So be−x=10y′′+9y. Also, subtract y′ from something? Alternatively, from y′=−3asin3x−be−x and y=acos3x+be−x, we can solve for acos3x and asin3x but it's messy.
12. Instead, compute the target expression directly using y′′ and then substitute a and b from y and y′
We have:
y′′(3sin3x−cos3x)=(−9acos3x+be−x)(3sin3x−cos3x)
Expand:
=−27acos3xsin3x+9acos23x+3be−xsin3x−be−xcos3x
Now, note that y′=−3asin3x−be−x, so −3asin3x=y′+be−x. Multiply by 9cos3x:
−27acos3xsin3x=9cos3x(y′+be−x)=9y′cos3x+9be−xcos3x
So the expression becomes:
=9y′cos3x+9be−xcos3x+9acos23x+3be−xsin3x−be−xcos3x
Combine be−xcos3x: 9−1=8, so:
=9y′cos3x+9acos23x+3be−xsin3x+8be−xcos3x
Now, 9acos23x=9cos3x(acos3x)=9cos3x(y−be−x)=9ycos3x−9be−xcos3x.
Substitute:
=9y′cos3x+(9ycos3x−9be−xcos3x)+3be−xsin3x+8be−xcos3x
=9y′cos3x+9ycos3x+3be−xsin3x−be−xcos3x
13. Now write be−x in terms of y and y′
From y=acos3x+be−x and y′=−3asin3x−be−x, we can eliminate a: multiply y by 3sin3x and y′ by cos3x? Better: solve for be−x by adding 3sin3x⋅y and cos3x⋅y′? Let's try:
3ysin3x=3acos3xsin3x+3be−xsin3x
y′cos3x=−3asin3xcos3x−be−xcos3x
Add them:
3ysin3x+y′cos3x=3be−xsin3x−be−xcos3x
That's exactly the leftover terms! So:
3be−xsin3x−be−xcos3x=3ysin3x+y′cos3x
14. Substitute back
Our expression becomes:
=9y′cos3x+9ycos3x+(3ysin3x+y′cos3x)
=(9y′cos3x+y′cos3x)+9ycos3x+3ysin3x
=10y′cos3x+9ycos3x+3ysin3x
=10y′cos3x+3y(sin3x+3cos3x)
This matches option (B) exactly.
Watch outA common mistake is to try to solve for a and b explicitly, which leads to messy algebra. The elegant path is to recognize that 3ysin3x+y′cos3x simplifies the be−x terms directly.
TipThe combination 3ysin3x+y′cos3x appears naturally when you add 3sin3x times y and cos3x times y′ — a neat trick to eliminate a and express everything in terms of y and y′.
✓Final answerThe correct option is (B).
ANSWER: B
- TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQQ.If (a+bx)exy=x, then dx2d2y= (A) x31(xy′+y2)2 (B) x31(xy′+y2) (C) x31(xy′−y) (D) x31(xy′−y)2
›Reveal solutionSolution
By simplifying the given equation using logarithms and then applying implicit differentiation twice, we find that the second derivative dx2d2y is x31(xy′−y)2.
The problem asks us to find the second derivative dx2d2y from the given implicit relation (a+bx)exy=x. The presence of the exponential term exy suggests that taking the natural logarithm might simplify the expression, making differentiation easier. The options provided involve the term (xy′−y), which is a strong hint that we should try to express our derivatives in terms of this quantity.
Here's a step-by-step approach:
-
Simplify the given equation:
The initial equation is (a+bx)exy=x.
To simplify, first isolate the exponential term:
exy=a+bxx
Now, take the natural logarithm on both sides. This brings the exponent down, making the equation linear in xy:
log(exy)=log(a+bxx)
Using the logarithm property log(A/B)=logA−logB:
xy=logx−log(a+bx)
Multiplying by x gives us an explicit expression for y:
y=x(logx−log(a+bx))
-
Find the first derivative, y′:
We differentiate y=x(logx−log(a+bx)) with respect to x. We will use the product rule, (uv)′=u′v+uv′.
Let u=x, so u′=1.
Let v=logx−log(a+bx). To find v′, we differentiate term by term:
dxd(logx)=x1
dxd(log(a+bx))=a+bx1⋅dxd(a+bx)=a+bxb
So, v′=x1−a+bxb.
Applying the product rule for y′:
y′=(1)⋅(logx−log(a+bx))+x⋅(x1−a+bxb)
y′=(logx−log(a+bx))+1−a+bxbx
From Step 1, we know that logx−log(a+bx)=xy. Substitute this back into the expression for y′:
y′=xy+1−a+bxbx
Combine the constant and fractional terms:
y′=xy+a+bx(a+bx)−bx
y′=xy+a+bxa
-
Express xy′−y:
The options involve the term (xy′−y). Let's rearrange our expression for y′ from Step 2 to find this:
y′−xy=a+bxa
Multiply the entire equation by x:
x(y′−xy)=x(a+bxa)
xy′−y=a+bxax
This is a key intermediate result.
-
Find the second derivative, y′′:
Now we differentiate y′=xy+a+bxa with respect to x to find y′′.
y′′=dxd(xy)+dxd(a+bxa)
For the first term, dxd(xy), use the quotient rule (vu)′=v2u′v−uv′:
dxd(xy)=x2y′⋅x−y⋅1=x2xy′−y
For the second term, dxd(a+bxa), treat it as a(a+bx)−1 and use the chain rule:
dxd(a(a+bx)−1)=a⋅(−1)(a+bx)−2⋅dxd(a+bx)
=−a(a+bx)−2⋅b=−(a+bx)2ab
Combining these, we get y′′:
y′′=x2xy′−y−(a+bx)2ab
-
Substitute and simplify to match the options:
We have the expression for y′′ in terms of xy′−y, x, a, and b. Now, substitute the result from Step 3, xy′−y=a+bxax, into the y′′ expression:
y′′=x2a+bxax−(a+bx)2ab
Simplify the first term:
y′′=x2(a+bx)ax−(a+bx)2ab
y′′=x(a+bx)a−(a+bx)2ab
To combine these fractions, find a common denominator, which is x(a+bx)2:
y′′=x(a+bx)2a(a+bx)−x(a+bx)2abx
y′′=x(a+bx)2a2+abx−abx
y′′=x(a+bx)2a2
Finally, we need to express this result in terms of (xy′−y) to match the given options.
From Step 3, we know xy′−y=a+bxax.
We can rearrange this to isolate a+bxa:
a+bxa=xxy′−y
Now, square both sides:
(a+bxa)2=(xxy′−y)2=x2(xy′−y)2
Observe that our expression for y′′ can be written as:
y′′=x1⋅(a+bx)2a2=x1⋅(a+bxa)2
Substitute the squared expression back into y′′:
y′′=x1⋅x2(xy′−y)2
y′′=x3(xy′−y)2
The final expression matches option (D).
✓Final answerThe second derivative is x31(xy′−y)2.
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- TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQQ.If 2x2+3xy−y2+4x−5y+6=0, then the value of dxdy at (x,y)=(1,−2) is (A) 1 (B) −1 (C) 27 (D) 0
›Reveal solutionSolution
Use implicit differentiation on the given polynomial, then substitute the point (1, –2) to solve for dy/dx. The result is 0, so option (D) is correct.
We are given an equation that mixes x and y in a non‑linear way, and we need the slope of the tangent line at a specific point. Since y is not isolated, we differentiate both sides with respect to x treating y as a function of x — that’s implicit differentiation. The key idea: every time we differentiate a term with y, we multiply by dxdy (chain rule). Then we plug in the coordinates to get a numerical value.
- Differentiate term by term Start with
2x2+3xy−y2+4x−5y+6=0.
Differentiate each term with respect to x:
- dxd(2x2)=4x
- dxd(3xy): use product rule — 3⋅(1⋅y+x⋅dxdy)=3y+3xdxdy
- dxd(−y2)=−2ydxdy
- dxd(4x)=4
- dxd(−5y)=−5dxdy
- dxd(6)=0
- Collect the derivative terms Putting it all together:
4x+3y+3xdxdy−2ydxdy+4−5dxdy=0.
Group the terms containing dxdy:
(3x−2y−5)dxdy+(4x+3y+4)=0.
- Solve for dxdy
(3x−2y−5)dxdy=−(4x+3y+4)
dxdy=−3x−2y−54x+3y+4.
- Substitute the point (x,y)=(1,−2) Numerator: 4(1)+3(−2)+4=4−6+4=2 Denominator: 3(1)−2(−2)−5=3+4−5=2 Hence
dxdy=−22=−1.
Watch outA common mistake is forgetting the minus sign when moving terms, or misapplying the product rule on 3xy. Always check that each y-term gets a dxdy factor.
TipYou can verify by solving the quadratic for y at x=1 and checking the slope of the tangent — but implicit differentiation is faster and avoids messy algebra.
✓Final answerThe correct option is (B).
ANSWER: B
- TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQQ.Derivative of (sinx)x with respect to x(sinx) is (A) x(sinx)[xcosx(logx)+sinx](sinx)x−1[(sinx)log(sinx)+xcosx] (B) x(sinx)[xcosx(logx)+sinx](sinx)x[(sinx)(log(sinx))+xcosx] (C) (sinx)x−1[(sinx)log(sinx)+xcosx]xsinx−1[xcosx(logx)+sinx] (D) (sinx)x[(sinx)log(sinx)+xcosx]xsinx[xcosx(logx)+sinx]
›Reveal solutionSolution
Differentiate both u=(sinx)x and v=xsinx logarithmically, then form dv/dxdu/dx. The numerator is (sinx)x−1[sinxlog(sinx)+xcosx] and the denominator carries [xcosxlogx+sinx] — option (A).
The concept first. Two ideas combine.
- Derivative of one function w.r.t. another. By the chain rule, dvdu=dv/dxdu/dx. So we never need to eliminate x; we just differentiate each separately and divide.
- Logarithmic differentiation. Neither the power rule (xn, constant exponent) nor the exponential rule (ax, constant base) applies when both base and exponent depend on x. Taking log first turns the exponent into a product, which the product rule can handle.
Step 1 — Differentiate u=(sinx)x.
logu=xlog(sinx)
Differentiate both sides:
u1dxdu=log(sinx)+x⋅sinxcosx
dxdu=(sinx)x[log(sinx)+sinxxcosx]=(sinx)x−1[sinxlog(sinx)+xcosx]
(the last step just takes one factor of sinx out of the bracket).
Step 2 — Differentiate v=xsinx.
logv=sinxlogx
v1dxdv=cosxlogx+xsinx
dxdv=xsinx[cosxlogx+xsinx]=xsinx⋅x1[xcosxlogx+sinx]
Step 3 — Divide.
dvdu=dv/dxdu/dx=xsinx[xcosxlogx+sinx](sinx)x−1[sinxlog(sinx)+xcosx]
Step 4 — Recognise the shape. The (sinx)-power sits on top (because u is the function being differentiated) and the x-power sits below, with the bracket [sinxlog(sinx)+xcosx] belonging to u and [xcosxlogx+sinx] belonging to v. Options (C) and (D) invert this — they give dv/du, the derivative the wrong way round.
✓Final answerThe derivative is xsinx[xcosxlogx+sinx](sinx)x−1[sinxlog(sinx)+xcosx], so the correct option is (A).
ANSWER: A
- TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQQ.The differential equation for which y2=4a(x+a) (where a is a parameter) is general solution, is (A) y−yy′2=2xy′ (B) y+yy′2=2xy′ (C) y(y+y′)=2xy′ (D) y(y−y′)=2xy′
›Reveal solutionSolution
To find the differential equation, we eliminate the arbitrary parameter a from the given general solution y2=4a(x+a) by differentiating it once. The resulting differential equation is y−yy′2=2xy′.
When we are given a general solution to a differential equation, it contains one or more arbitrary constants (parameters). The process of finding the differential equation involves eliminating these arbitrary constants. The number of times we need to differentiate the given solution is equal to the number of independent arbitrary constants present in it.
In this problem, the general solution is y2=4a(x+a), and a is the single arbitrary parameter. Therefore, we will differentiate the equation once with respect to x to obtain an expression involving a, and then substitute this expression back into the original equation to eliminate a.
Here's the step-by-step derivation:
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Identify the arbitrary parameter:
The given general solution is y2=4a(x+a).
Here, a is the only arbitrary parameter. Since there is one arbitrary parameter, the differential equation will be of the first order, meaning we will need to differentiate the given equation once with respect to x.
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Differentiate the equation with respect to x:
We differentiate both sides of y2=4a(x+a) with respect to x. Remember to use the chain rule for y2 and treat a as a constant.
dxd(y2)=dxd(4a(x+a))
2ydxdy=4adxd(x+a)
2yy′=4a(1+0)
2yy′=4a
From this, we can express $a$ in terms of $y$ and $y'$:a=42yy′
a=2yy′
- Eliminate the parameter a: Now, substitute the expression for a back into the original general solution y2=4a(x+a).
y2=4(2yy′)(x+2yy′)
Simplify the equation:y2=2yy′(x+2yy′)
Assuming $y \neq 0$, we can divide both sides by $y$:y=2y′(x+2yy′)
Now, distribute $2y'$ on the right side:y=2xy′+2y′(2yy′)
y=2xy′+y(y′)2
Rearrange the terms to match the given options:y−y(y′)2=2xy′
This can also be written as:y−yy′2=2xy′
- Compare with the options: Comparing our derived differential equation y−yy′2=2xy′ with the given options: (A) y−yy′2=2xy′ (B) y+yy′2=2xy′ (C) y(y+y′)=2xy′ (D) y(y−y′)=2xy′ Our result matches option (A).
✓Final answerThe differential equation for which y2=4a(x+a) is the general solution is (A)y−yy′2=2xy′.
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- TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQQ.If the perpendicular distance from the focus of an ellipse 9x2+b2y2=1 (b<3) to its corresponding directrix is 54, then the slope of the tangent to this ellipse drawn at (23,2b) is (A) −32 (B) 32 (C) 23 (D) −23
›Reveal solutionSolution
e=35,b2=4; the tangent slope at the given point is −32.
For 9x2+b2y2=1 with a=3, the focus-to-directrix distance is
ea−ae=ea(1−e2)=54.
With a=3: 35e2+4e−35=0⇒e=35.
Then e2=95 and b2=a2(1−e2)=9(94)=4, so the ellipse is 9x2+4y2=1.
The evaluation point is (23,2b)=(23,2).
Differentiating implicitly: 92x+2yy′=0⇒y′=−9y4x.
y′=−924⋅23=−9⋅212=−32.
✓Final answerANSWER: A — slope =−32.
- TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQQ.If x2+xy+y2=k, then dx2d2y= (A) (x+2y)3−6k (B) (x+2y)2−6k (C) (2x+y)2x2+xy+y2 (D) 0
›Reveal solutionSolution
Implicit differentiation twice gives dx2d2y=(x+2y)3−6k — option (A).
Step 1 — First derivative. Differentiate x2+xy+y2=k:
2x+y+xy′+2yy′=0⇒y′=−x+2y2x+y.
Step 2 — Second derivative. With y′=−vu where u=2x+y, v=x+2y (so u′=2+y′, v′=1+2y′):
y′′=−v2u′v−uv′.
Compute the numerator:
u′v−uv′=(2+y′)(x+2y)−(2x+y)(1+2y′)=3y−3xy′.
Substitute y′=−x+2y2x+y:
3y−3x(−x+2y2x+y)=x+2y3y(x+2y)+3x(2x+y)=x+2y6(x2+xy+y2)=x+2y6k.
Step 3 — Assemble.
y′′=−v21⋅x+2y6k=−(x+2y)36k.
(Here x2+xy+y2=k was used.)
✓Final answerdx2d2y=(x+2y)3−6k — option (A).
- TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQQ.A function f:R→R is such that yf(x+y)+cosmy=1+yf(x). If m=2, then f′(x)= (A) −2sin2xy (B) 4x (C) y2sin2xy (D) 2x2
›Reveal solutionSolution
Rearranging the relation gives yf(x+y)−f(x)=y21−cos(mxy); letting y→0 and using 1−cosθ→θ2/2 yields f′(x)=2m2x2=2x2 for m=2 — option (D).
The concept first. The derivative is defined by
f′(x)=limy→0yf(x+y)−f(x)
So whenever a problem hands you a relation connecting f(x+y) and f(x), the strategy is always the same: isolate f(x+y)−f(x), divide by y, and take the limit. The answer must be a function of x alone — y is the vanishing increment, so any option still containing y (like (A) and (C)) cannot be a derivative at all.
Step 1 — Isolate the difference.
yf(x+y)+cos(mxy)=1+yf(x)
⇒yf(x+y)−yf(x)=1−cos(mxy)
⇒y[f(x+y)−f(x)]=1−cos(mxy)
Step 2 — Build the difference quotient. Divide both sides by y2 (valid for y=0):
yf(x+y)−f(x)=y21−cos(mxy)
Step 3 — Take the limit y→0.
Use the standard limit 1−cosθ=2sin2(2θ), so for small θ, 1−cosθ≈2θ2. With θ=mxy:
y21−cos(mxy)=y22sin2(2mxy)=2⋅(2mxysin2mxy)2⋅4m2x2y→02m2x2
Therefore
f′(x)=2m2x2
Step 4 — Put m=2.
f′(x)=222x2=24x2=2x2
Step 5 — Sanity check the options. (A) and (C) still contain y, which is impossible for f′(x). (B) 4x would arise from lim(1−cos(mxy))/y2 done wrongly. Only 2x2 survives.
✓Final answerf′(x)=2x2, so the correct option is (D).
ANSWER: D
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