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).

All Differentiation notes MHT-CET Maths strategy

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

Compute $\dfrac{du}{dx}$ and $\dfrac{dv}{dx}$ separately.
Then $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$ , provided $\dfrac{dv}{dx} \neq 0$ .

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

Derivative of u with respect to v

\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 \neq 0ratio is undefined where the bottom's derivative vanishes

Worked example

Find the derivative of

x^3

with respect to

x^2

Let $u = x^3$ and $v = x^2$ . These are differentiated with respect to $x$ , not each other.
Differentiate each with respect to $x$ : $\dfrac{du}{dx} = 3x^2$ and $\dfrac{dv}{dx} = 2x$ .
Form the ratio: $\dfrac{du}{dv} = \dfrac{3x^2}{2x} = \dfrac{3x}{2}$ .

Answer:

\dfrac{du}{dv} = \dfrac{3x}{2}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the derivative of

\log x

with respect to

x^2

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Derivative of $x^4$ with respect to $x^2$ .
2.
Derivative of $\sin x$ with respect to $\cos x$ .
3.
Derivative of $e^{2x}$ with respect to $e^x$ .
4.
Derivative of $x^2$ with respect to $\dfrac{1}{x}$ .

From the bank · past-year question

Example 1DifferentiationMODERATE

Derivative of

e^{\sqrt{x}}

w.r.t.

\sqrt{x}

[Q124 · 9th May Shift 1 · 2023]

Do NOT differentiate one function directly by the other

\dfrac{du}{dv}

is not 'differentiate

u

and substitute

v

'. You must form both

\dfrac{du}{dx}

and

\dfrac{dv}{dx}

and divide. There is no shortcut that skips

x

Substitute the point only after dividing

If a value like

x = 5

is given, keep

x

symbolic while you form

\dfrac{du/dx}{dv/dx}

, then substitute. Plugging the point into

\dfrac{du}{dx}

and

\dfrac{dv}{dx}

before simplifying invites arithmetic slips.

The bottom's derivative must be non-zero

\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}

is only valid where

\dfrac{dv}{dx} \neq 0

. If

v = g(x)

has a stationary point at the value asked, the rate of change with respect to

v

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)

and

g(\tan x)

, and hands you the values of

f'

and

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'

g'

value both ride along.

Definition

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

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

Composite-over-composite ratio

\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)

with respect to

g(\cos x)

x = \dfrac{\pi}{3}

, given

f'\!\left(\dfrac{\sqrt{3}}{2}\right) = 4

and

g'\!\left(\dfrac{1}{2}\right) = 2

Chain-rule the top: $\dfrac{d}{dx}f(\sin x) = f'(\sin x)\cdot\cos x$ .
Chain-rule the bottom: $\dfrac{d}{dx}g(\cos x) = g'(\cos x)\cdot(-\sin x)$ .
Form the ratio: $\dfrac{f'(\sin x)\,\cos x}{g'(\cos x)\,(-\sin x)}$ .
At $x = \dfrac{\pi}{3}$ : $\sin x = \dfrac{\sqrt{3}}{2}$ , $\cos x = \dfrac{1}{2}$ . Substitute the given values: $\dfrac{4 \cdot \frac{1}{2}}{2 \cdot \left(-\frac{\sqrt{3}}{2}\right)} = \dfrac{2}{-\sqrt{3}}$ .

Answer:

-\dfrac{2}{\sqrt{3}}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the derivative of

f(x^2)

with respect to

g(x^3)

x = 1

, given

f'(1) = 6

and

g'(1) = 2

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Derivative of $f(x^2)$ w.r.t. $g(x)$ at $x = 2$ , given $f'(4) = 3$ , $g'(2) = 1$ .
2.
Derivative of $f(\log x)$ w.r.t. $g(x)$ at $x = 1$ , given $f'(0) = 5$ , $g'(1) = 1$ .
3.
Derivative of $f(\sin x)$ w.r.t. $g(\sin x)$ at any $x$ , given $f'(t) = 3$ , $g'(t) = 2$ for all $t$ .
4.
Derivative of $f(e^x)$ w.r.t. $g(x)$ at $x = 0$ , given $f'(1) = 7$ , $g'(0) = 7$ .

From the bank · past-year question

Example 2DifferentiationHARD

The derivative of

f(\sec x)

with respect to

g(\tan x)

x=\frac{\pi}{4}

, where

f'(2)=4

and

g'(1)=2

, is

[Q105 · 9th May Shift 1 · 2024]

Each inner derivative must be carried through

When the top is

f(\sec x)

, its derivative is

f'(\sec x)\cdot\sec x\tan x

— the inner

\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'(\sqrt{2})

is meant for where the inner function equals

\sqrt{2}

(e.g.

\sec\frac{\pi}{4} = \sqrt{2}

) — not for

x = \sqrt{2}

. Evaluate the inner function at the given

x

first, then read off the matching

f'

value.

Keep the negative sign on falling inner functions

If the bottom is

g(\cos x)

, its inner derivative is

-\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'

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Differentiating One Function With Respect to Another
Derivative of u with respect to v
$\dfrac{du}{dv} = \dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}, \qquad \dfrac{dv}{dx} \neq 0$
Composite Functions Using Given Derivatives f' and g'
Composite-over-composite ratio
$\dfrac{d\,[f(p(x))]}{d\,[g(q(x))]} = \dfrac{f'(p(x))\,p'(x)}{g'(q(x))\,q'(x)}$

Watch out for (6)

Do NOT differentiate one function directly by the other
→ Differentiating One Function With Respect to Another
Substitute the point only after dividing
→ Differentiating One Function With Respect to Another
The bottom's derivative must be non-zero
→ Differentiating One Function With Respect to Another
Each inner derivative must be carried through
→ Composite Functions Using Given Derivatives f' and g'
Match each supplied value to the right inner argument
→ Composite Functions Using Given Derivatives f' and g'
Keep the negative sign on falling inner functions
→ Composite Functions Using Given Derivatives f' and g'

Mastery check — 3 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1DifferentiationHARD

The rate of change of

\sqrt{x^2+16}

with respect to

\frac{x}{x-1}

x=5

[Q127 · 9th May Shift 1 · 2024]

Example 2DifferentiationHARD

The derivative of

f(\tan x)

w.r.t.

g(\sec x)

x=\frac{\pi}{4}

where

f'(1)=2

and

g'(\sqrt{2})

[Q109 · 12th May Shift 1 · 2024]

Example 3DifferentiationMODERATE

Derivative of

\sin^{2}x

with respect to

e^{\cos x}

[Q146 · 3rd May Shift 2 · 2023]

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