Q.Differentiate w.r.t. x, the following function:
Concept understanding — Chain Rule
The Chain Rule: Why It Makes Sense
Imagine you're assembling a toy. First you put part A into part B, then you put that combined piece into part C. The final toy's position depends on how you moved A, which then affected B, which then affected C. That's exactly what the chain rule captures — how a change in the first variable ripples through a sequence of functions to affect the final output.
Let's make this concrete. Suppose you have a function f that depends on g, and g itself depends on x:
y=f(g(x))
You want to know: if x changes by a tiny amount, how much does y change? The answer isn't just f′(g(x)) — because g(x) itself changes when x changes. You have to multiply the two rates:
- How fast does g change with respect to x? That's g′(x).
- How fast does f change with respect to its input g? That's f′(g(x)).
The total effect is the product:
dxdy=f′(g(x))⋅g′(x)
In Leibniz notation, this looks even more natural: dxdy=dudy⋅dxdu, where u=g(x). The du's "cancel" like fractions — though this is just a helpful memory aid, not a rigorous proof.
The Precise Statement
Chain Rule (single variable): If g is differentiable at x and f is differentiable at g(x), then the composite function h(x)=f(g(x)) is differentiable at x, and
h′(x)=f′(g(x))⋅g′(x)
That's it. One multiplication. But the power is enormous — it lets you differentiate almost any nested function.
A Simple Example
Differentiate h(x)=sin(3x2).
Here f(u)=sinu and g(x)=3x2. Then:
- f′(u)=cosu, so f′(g(x))=cos(3x2)
- g′(x)=6x
Multiply: h′(x)=cos(3x2)⋅6x=6xcos(3x2)
The most common mistake is forgetting to multiply by the inner derivative. Students often write dxdsin(3x2)=cos(3x2) and stop — that's wrong. The chain rule demands you also multiply by 6x.
Why It's Called a "Chain"
Think of a chain of links: x→g→f. Each link has its own rate of change. To find the total rate from x to f, you multiply the rates of each link. If you had three functions — say h(x)=f(g(k(x))) — you'd multiply three derivatives:
h′(x)=f′(g(k(x)))⋅g′(k(x))⋅k′(x)
The chain can be as long as you like. Each new function adds one more factor.
The Intuition in One Sentence
The chain rule says: the rate of change of the whole is the product of the rates of change of the parts, evaluated at the right places.
The chain rule is not optional — it's the backbone of calculus. Every derivative of a trigonometric, exponential, logarithmic, or power function that isn't just xn uses it. Master this, and you master differentiation.
The chain rule is one of the most heavily tested formulas in the NCERT Class 12 Continuity and Differentiability chapter, and it underlies nearly every differentiation problem in CBSE boards, JEE Main and JEE Advanced. Whether you're searching 'chain rule differentiation class 12 examples' or 'chain rule important questions for JEE', this f'(g(x))·g'(x) pattern is the formula every subsequent derivative rule in the syllabus builds on.
- Write each term as a power and use the chain rule. 3x+2=(3x+2)1/2: derivative =21(3x+2)−1/2⋅3=23x+23. 2x2+41=(2x2+4)−1/2: derivative =−21(2x2+4)−3/2⋅4x=−(2x2+4)3/22x. Adding, the derivative is 23x+23−(2x2+4)3/22x.
- With the NCERT convention logx=lnx, use dxdlogau=ulna1⋅u′. Here u=logx, u′=x1:
dxdlog7(logx)=(logx)ln71⋅x1=xln7logx1.
✓Final answer- 23x+23−(2x2+4)3/22x;
- xln7logx1 (with logx=lnx)
Differentiating term-by-term with the chain rule: (i) 23x+23−(2x2+4)3/22x;
(ii) xln7logx1 (taking logx=lnx, the NCERT convention).
(i) 3x+2+2x2+41
Each term is an outer power wrapped around an inner expression, so we use dxd[g(x)]n=n[g(x)]n−1g′(x).
First term: (3x+2)1/2.
dxd(3x+2)1/2=21(3x+2)−1/2⋅3=23x+23.
Second term: (2x2+4)−1/2.
dxd(2x2+4)−1/2=−21(2x2+4)−3/2⋅4x=−(2x2+4)3/22x.
Sum of the derivatives:
dxdy=23x+23−(2x2+4)3/22x.
The reciprocal square root carries a negative power, so its derivative is negative — keep that minus sign.
(ii) log7(logx)
Use the base-a log rule dxdlogau=ulna1⋅dxdu, with the standard NCERT convention that logx denotes the natural logarithm lnx (so dxdlogx=x1).
Here the base is a=7 and u=logx:
dxdlog7(logx)=(logx)ln71⋅dxd(logx)=(logx)ln71⋅x1.
Therefore
dxdlog7(logx)=xln7logx1.
dxdlogau=ulna1⋅dxdu
- 23x+23−(2x2+4)3/22x;
- xln7logx1 (with logx=lnx)
Method: Differentiating Radical and Logarithmic Composite Functions
This method applies to differentiating expressions built from roots (fractional powers) and logarithms of a base other than e, where an inner function is nested inside an outer power or log.
Steps
Step 1: Rewrite radicals as fractional powers
Convert every ⋅ or ⋅1 into (⋅)1/2 or (⋅)−1/2 so the ordinary power rule can be applied directly.
Step 2: Apply the chain rule with the power rule (for radicals) or the base-change log rule (for logs)
For a radical term [g(x)]n:
dxd[g(x)]n=n[g(x)]n−1g′(x).
For a log-of-a-log expression loga(u(x)), use the base-a rule (which reduces to the natural-log derivative scaled by a constant):
dxdlogau=ulna1⋅dxdu.
Step 3: Combine the pieces and simplify
Substitute the inner function's own derivative g′(x) or u′(x), multiply through, and simplify any resulting fraction — combine terms over a common denominator only if the question asks for a single expression.
Common Mistakes
Mistake 1: Dropping the negative sign on a reciprocal-power (negative-exponent) term
Why it's wrong: writing 2x2+41=(2x2+4)−1/2 and then differentiating without carrying the negative exponent through the power rule gives a positive derivative where a negative one is required. Correct approach: keep the exponent explicitly negative throughout — n=−21 — so the power rule's n[g(x)]n−1 naturally produces the correct sign.
Mistake 2: Forgetting the outer lna1 factor when differentiating loga(u)
Why it's wrong: differentiating log7(logx) as if it were log(logx) (base e) and skipping the ln7 in the denominator gives an answer too large by a factor of ln7. Correct approach: always convert logau to ulna1⋅u′ explicitly, writing the base's natural log in the denominator before simplifying anything else.
- GUJCET 2026Set x1 markMCQQ.If y=log2026(log2025x), then dxdy= ______ (A) xlogxlog20251 (B) xlogxlog20261 (C) 2025xlogx1 (D) 2026xlogx1
›Reveal solutionSolution
Convert the base-changed logs to natural logs and apply the chain rule.
Write y=ln2026ln(log2025x) where u=log2025x=ln2025lnx.
dxdy=ln20261⋅u1⋅dxdu,dxdu=ln20251⋅x1.
Substituting u=ln2025lnx:
dxdy=ln20261⋅lnxln2025⋅xln20251=xlnxln20261.
✓Final answerxlogxlog20261
ANSWER: (B)
- GUJCET 2021Set 151 markMCQQ.dxd(cosec−1ex)=. (A) e2x−11 (B) e2x−1−1 (C) sin−1(ex) (D) e2x−1−ex
›Reveal solutionSolution
Apply the chain rule to cosec−1u with u=ex.
Concept: dxdcosec−1u=∣u∣u2−1−1dxdu. With u=ex, u′=ex:
exe2x−1−ex=e2x−1−1.
✓Final answer(B) e2x−1−1
ANSWER: (B)
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.dxd(sin(log7x)) = ____ ; (x>0)(a) xcos(log7x)(b) xlog7cos(log7x)(c) log7cos(logx)(d) xcos(logx)
›Reveal solutionSolution
Convert log7x to natural log form before applying the chain rule.
log7x=log7logx, so dxd(log7x)=xlog71.
By the chain rule, dxdsin(log7x)=cos(log7x)⋅xlog71=xlog7cos(log7x).
✓Final answerThe correct option is (b) xlog7cos(log7x).
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.What is differentiation of cos−1(sinx) with respect to x?(a) 1(b) −1(c) 2π−1(d) 2π
›Reveal solutionSolution
Rewrite cos−1(sinx) as 2π−x, then differentiate.
Using sinx=cos(2π−x), we get cos−1(sinx)=2π−x on the interval where 2π−x∈[0,π].
dxd(2π−x)=−1.
(Directly: dxdcos−1(sinx)=1−sin2x−cosx=∣cosx∣−cosx=−1 where cosx>0.)
✓Final answer(b) −1.
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.dxd(xsinx)= ___, 0<x<π.(a) 2xsinx1(b) 2xsinxxcosx(c) 2xsinxxcosx+sinx(d) xsinxxsinx+cosx
›Reveal solutionSolution
Differentiate u with u=xsinx via the chain and product rules.
Let u=xsinx, so dxdu=2uu′.
By the product rule u′=sinx+xcosx.
Hence
dxdxsinx=2xsinxxcosx+sinx.
✓Final answer(c) 2xsinxxcosx+sinx.
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