Q.Find the second order derivative of the function: x20
Concept understanding — Successive Differentiation
Successive Differentiation — Repeated Slopes
Differentiate a function, then differentiate the result, then differentiate that, and so on. Each pass produces a new function describing a deeper layer of change. The physics picture makes it concrete: position → velocity (1st derivative) → acceleration (2nd) → jerk (3rd). Every step asks the same question: "how does the previous rate of change itself change?"
The definition and notation
If y=f(x), its successive derivatives are written y1,y2,…,yn, equivalently f′(x),f′′(x),…,f(n)(x) or dxdy,dx2d2y,…,dxndny. Each one is the derivative of the previous:
dxndny=dxd(dxn−1dn−1y).
| Order | Leibniz | Lagrange | Newton |
|---|---|---|---|
| 1st | dxdy | f′(x) | y˙ |
| 2nd | dx2d2y | f′′(x) | y¨ |
| nth | dxndny | f(n)(x) | — |
Seeing the pattern
Take y=x4: y1=4x3,y2=12x2,y3=24x,y4=24,y5=0. Each differentiation drops the degree by one, so a degree-n polynomial has a constant nth derivative and vanishing higher ones. Other families behave differently: eax never dies out (dxndneax=aneax), and sinx cycles every four steps (sin→cos→−sin→−cos).
Power rule applied n times: dxndn(xm)=m(m−1)⋯(m−n+1)xm−n for n≤m.
dx2d2y is not (dxdy)2 — a second derivative is not the square of the first derivative.
Why it matters
Higher derivatives power the second-derivative test for maxima and minima, Taylor and Maclaurin expansions, Leibniz's theorem for the nth derivative of a product, and differential equations such as F=ma (a second derivative of position).
Successive (higher-order) differentiation is its own named section in the NCERT Class 12 Continuity and Differentiability chapter, and finding a general nth-derivative pattern for polynomials, exponentials or sine is a recurring CBSE board and JEE Main question type. Students searching 'successive differentiation class 12 examples' or 'nth derivative formula' will recognize this repeated-differentiation notation (y₁, y₂, ..., yₙ) as the standard exam convention.
Differentiate x20 twice using the power rule dxdxn=nxn−1 — this is successive (repeated) differentiation.
First derivative:
dxdy=20x19
Second derivative — differentiate the result again:
dx2d2y=20⋅19x18=380x18
dx2d2y=380x18
Applying the power rule twice to x20 gives dx2d2y=380x18.
A second-order derivative just means we differentiate, then differentiate the result again. For a pure power of x the only tool we need is the power rule, dxdxn=nxn−1 — no chain rule, no product rule.
Step 1 — First derivative
For y=x20,
dxdy=20x20−1=20x19.
Step 2 — Second derivative
Now differentiate 20x19. The constant 20 stays put; apply the power rule to x19:
dx2d2y=20⋅19x19−1=20⋅19x18.
Step 3 — Simplify
Since 20×19=380,
dx2d2y=380x18.
In one shot, dx2d2xn=n(n−1)xn−2. With n=20: 20⋅19=380 and the exponent drops to 18.
dx2d2y=380x18
Method: Successive Differentiation of a Power Function
This method finds a second (or higher) order derivative of a pure power xn by applying the power rule repeatedly, one order at a time.
Steps
Step 1: Identify the function type and the order of derivative required
Check whether the expression is a pure power of x (possibly with a constant coefficient). For a pure power, no chain rule, product rule, or quotient rule is needed — only the power rule, applied as many times as the required order.
dxd(xn)=nxn−1
Step 2: Differentiate once to get the first derivative
Apply the power rule to the original function to obtain y1=dxdy. This reduces the exponent by 1 and multiplies by the original exponent.
Step 3: Differentiate the result again for the second derivative
Treat y1 as a new function and apply the power rule to it directly — differentiate the coefficient-power expression, not the original function. This gives y2=dx2d2y.
Step 4: Simplify the constant multiplier
Multiply out any numerical coefficients that arise from repeated application (e.g., n(n−1)) and leave the answer as a single coefficient times a power of x. For higher orders, keep repeating Steps 2–3, tracking how the exponent decreases and the coefficient grows by successive multiplication.
Common Mistakes
Mistake 1: Stopping after computing only the first derivative
Why it's wrong: the question asks for the second order derivative, but 20x19 is only an intermediate step. Correct approach: always re-read what order is asked, and explicitly differentiate the first-derivative expression once more before writing the final answer.
Mistake 2: Forgetting to multiply the existing coefficient into the new one
Why it's wrong: when differentiating 20x19, both the coefficient 20 and the exponent 19 must be multiplied together (giving 380) — some students only bring down the exponent and forget to multiply it by the coefficient already present. Correct approach: apply the power rule to the whole term as 20⋅19x18, not just x18.
Mistake 3: Confusing dx2d2y with (dxdy)2
Why it's wrong: squaring the first derivative gives a completely different (and wrong) expression — the second derivative is the derivative of the derivative, not its square. Correct approach: always compute the second derivative by differentiating the first-derivative expression again, never by squaring it.
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.If ex(y+1)=1, then dx2d2y= ____.(a) y(b) y+1(c) dxdy(d) dxdy+1
›Reveal solutionSolution
Solve for y explicitly, differentiate twice, and express the result back in terms of y.
ex(y+1)=1⇒y+1=e−x⇒y=e−x−1.
dxdy=−e−x=−(y+1).
dx2d2y=e−x=y+1.
✓Final answerThe correct option is (b) y+1.
- GUJCET 2022Set 081 markMCQQ.If y=100e2x+200e−2x and dx2d2y=ay then a= ______. (A) 4 (B) −4 (C) 2 (D) 0
›Reveal solutionSolution
Differentiating twice each exponential brings down a factor 4, giving y′′=4y.
Concept. y=100e2x+200e−2x.
y′=200e2x−400e−2x, and y′′=400e2x+800e−2x=4(100e2x+200e−2x)=4y.
Hence a=4.
✓Final answer(A) 4
ANSWER: (A)
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.If ey(x+1)=1, then what is the value of dx2d2y?(a) −(dxdy)(b) (dxdy)(c) −(dxdy)2(d) (dxdy)2
›Reveal solutionSolution
Solve for y explicitly, then relate y′′ to (y′)2.
ey(x+1)=1⇒ey=x+11⇒y=−ln(x+1).
dxdy=−x+11,dx2d2y=(x+1)21.
Since (dxdy)2=(x+1)21, we get dx2d2y=(dxdy)2.
✓Final answer(d) (dxdy)2.
- GUJCET 2021Set 151 markMCQQ.If x+1=e−y, then dx2d2y=. (A) (dxdy)3 (B) dxdy (C) (dxdy)2 (D) −dxdy
›Reveal solutionSolution
Differentiate twice and recognize dx2d2y=(dxdy)2.
Concept: x+1=e−y⇒y=−log(x+1), so
dxdy=−x+11,dx2d2y=(x+1)21.
Since (dxdy)2=(x+1)21, we get dx2d2y=(dxdy)2.
✓Final answer(C) (dxdy)2
ANSWER: (C)
- GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQQ.If y=loge(logex), then dx2d2y= ___ (where x>1).(a) −(x⋅logex)2loge(ex)(b) (x⋅logex)2loge(ex)(c) −loge(ex)(x⋅logex)2(d) (x⋅logxx)2loge(e/x)
›Reveal solutionSolution
Differentiate y=ln(lnx) twice using the chain and quotient rules, then recognise 1+lnx=ln(ex).
y=ln(lnx). First derivative: y′=lnx1⋅x1=xlnx1.
Second derivative, using y′=(xlnx)−1: y′′=−(xlnx)−2⋅dxd(xlnx).
dxd(xlnx)=lnx+1, so y′′=−(xlnx)2lnx+1.
Since 1+lnx=lne+lnx=ln(ex): y′′=−(xlogex)2loge(ex).
✓Final answer(a) −(xlogex)2loge(ex).
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.For the function y=tan−1x, (1+x2)y2= ___.(a) −2xy1(b) xy1(c) 2xy1(d) −xy1
›Reveal solutionSolution
Differentiate the first-order relation to get a relation for the second derivative.
For y=tan−1x, y1=1+x21, so (1+x2)y1=1.
Differentiate both sides:
(1+x2)y2+2xy1=0⇒(1+x2)y2=−2xy1.
✓Final answer(a) −2xy1.
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.