MHT-CET Maths · Differentiation

Differentiating One Function With Respect to Another

To find how u changes with respect to v (not x), differentiate both with respect to x and divide: du/dv equals (du/dx) over (dv/dx).

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Why this matters

This is a compact, formulaic technique that turns up almost every year as a quick scoring question — and it is pure mechanics once you see the trick. 5 PYQs sit directly here, 3 HARD and 2 MODERATE: most pair two composite functions and ask for their rate of change at a point, and the harder ones supply f' and g' at specific values so you must apply the chain rule to numerator and denominator separately. Master the single formula below and these become easy marks.

Concept 1 of 2

Differentiating One Function With Respect to Another

Intuition

When a question asks for the rate of change of one function with respect to ANOTHER function (not with respect to x), you cannot differentiate directly. The trick is to route both through x: differentiate the top function and the bottom function with respect to x, then divide. The shared dx cancels.

Definition

To differentiate u=f(x)u = f(x) with respect to v=g(x)v = g(x), differentiate each with respect to xx and take the ratio:

  • Compute dudx\dfrac{du}{dx} and dvdx\dfrac{dv}{dx} separately.
  • Then dudv=du/dxdv/dx\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}, provided dvdx≠0\dfrac{dv}{dx} \neq 0.

If a specific point is given, substitute it only after forming the ratio.

Derivative of u with respect to v

dudv=dudxdvdx,dvdx≠0\dfrac{du}{dv} = \dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}, \qquad \dfrac{dv}{dx} \neq 0
  • u = f(x)the function being differentiated (the 'top')
  • v = g(x)the function we differentiate with respect to (the 'bottom')
  • dv/dx≠0dv/dx \neq 0ratio is undefined where the bottom's derivative vanishes

Worked example

Find the derivative of x3x^3 with respect to x2x^2.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1DifferentiationMODERATE
Derivative of exe^{\sqrt{x}} w.r.t. x\sqrt{x} is

[Q124 · 9th May Shift 1 · 2023]

Do NOT differentiate one function directly by the other

dudv\dfrac{du}{dv} is not 'differentiate uu and substitute vv'. You must form both dudx\dfrac{du}{dx} and dvdx\dfrac{dv}{dx} and divide. There is no shortcut that skips xx.

Substitute the point only after dividing

If a value like x=5x = 5 is given, keep xx symbolic while you form du/dxdv/dx\dfrac{du/dx}{dv/dx}, then substitute. Plugging the point into dudx\dfrac{du}{dx} and dvdx\dfrac{dv}{dx} before simplifying invites arithmetic slips.

The bottom's derivative must be non-zero

dudv=du/dxdv/dx\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} is only valid where dvdx≠0\dfrac{dv}{dx} \neq 0. If v=g(x)v = g(x) has a stationary point at the value asked, the rate of change with respect to vv is undefined there.
Drill 2 more on differentiating one function with respect to another

Concept 2 of 2

Composite Functions Using Given Derivatives f' and g'

Intuition

A harder version pairs two composite functions like f(sec⁡x)f(\sec x) and g(tan⁡x)g(\tan x), and hands you the values of f′f' and g′g' at specific points. Same ratio idea — but now differentiating the top and bottom needs the chain rule, and the inner derivative and the supplied f′f'/g′g' value both ride along.

Definition

To differentiate f(p(x))f(p(x)) with respect to g(q(x))g(q(x)):

  • Chain-rule the top: ddxf(p(x))=f′(p(x))⋅p′(x)\dfrac{d}{dx}f(p(x)) = f'(p(x)) \cdot p'(x).
  • Chain-rule the bottom: ddxg(q(x))=g′(q(x))⋅q′(x)\dfrac{d}{dx}g(q(x)) = g'(q(x)) \cdot q'(x).
  • Take the ratio: f′(p(x)) p′(x)g′(q(x)) q′(x)\dfrac{f'(p(x))\,p'(x)}{g'(q(x))\,q'(x)}, then substitute the given point and the supplied values of f′f' and g′g'.

Composite-over-composite ratio

d [f(p(x))]d [g(q(x))]=f′(p(x)) p′(x)g′(q(x)) q′(x)\dfrac{d\,[f(p(x))]}{d\,[g(q(x))]} = \dfrac{f'(p(x))\,p'(x)}{g'(q(x))\,q'(x)}
  • f'(p(x))outer derivative of the top, read from the given f' value
  • p'(x)inner derivative of the top function
  • g'(q(x))outer derivative of the bottom, read from the given g' value
  • q'(x)inner derivative of the bottom function

Worked example

Find the derivative of f(sin⁡x)f(\sin x) with respect to g(cos⁡x)g(\cos x) at x=π3x = \dfrac{\pi}{3}, given f′ ⁣(32)=4f'\!\left(\dfrac{\sqrt{3}}{2}\right) = 4 and g′ ⁣(12)=2g'\!\left(\dfrac{1}{2}\right) = 2.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2DifferentiationHARD
The derivative of f(sec⁡x)f(\sec x) with respect to g(tan⁡x)g(\tan x) at x=π4x=\frac{\pi}{4}, where f′(2)=4f'(\sqrt{2})=4 and g′(1)=2g'(1)=2, is

[Q105 · 9th May Shift 1 · 2024]

Each inner derivative must be carried through

When the top is f(sec⁡x)f(\sec x), its derivative is f′(sec⁡x)⋅sec⁡xtan⁡xf'(\sec x)\cdot\sec x\tan x — the inner sec⁡xtan⁡x\sec x\tan x is part of it. Forgetting the inner derivative is the most common slip and silently drops a factor.

Match each supplied value to the right inner argument

A given f′(2)f'(\sqrt{2}) is meant for where the inner function equals 2\sqrt{2} (e.g. sec⁡π4=2\sec\frac{\pi}{4} = \sqrt{2}) — not for x=2x = \sqrt{2}. Evaluate the inner function at the given xx first, then read off the matching f′f' value.

Keep the negative sign on falling inner functions

If the bottom is g(cos⁡x)g(\cos x), its inner derivative is −sin⁡x-\sin x; the minus sign stays in the denominator and sets the sign of the final answer. Drop it and you get the right magnitude with the wrong sign.
Drill 1 more on composite functions using given derivatives f' and g'

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