MHT-CET Maths · Differentiation

Logarithmic Differentiation — Logs, Powers, and Long Products

When a function is a product, a quotient, or has a variable in the exponent, take the natural log of both sides FIRST — logs turn products into sums and pull exponents down front, so the differentiation becomes routine.

All Differentiation notes MHT-CET Maths strategy

Why this matters

This is the most mechanical high-yield method in the chapter — 32 PYQs sit here, 11 HARD, 19 MODERATE and 2 EASY, yet almost every one follows the SAME three steps. Two shapes dominate the exam: a variable raised to a variable power like (sin x) to the tan x, and a long product (x+1)(2x+1)…(nx+1) whose derivative is asked at x = 0. Master the three steps — take log, differentiate (1/y)·y′, multiply back by y — and most of these become one-minute questions.

Concept 1 of 5: Logarithmic Differentiation — the Method

y

is a product, a quotient, or — crucially — has a variable in BOTH the base and the exponent, the ordinary power rule and chain rule do not apply directly. Taking the natural log of both sides first turns multiplication into addition and pulls exponents down as multipliers, after which differentiation is straightforward.

Definition

The three-step method for $y = f(x)$ :

Take logs: write $\log y = \log f(x)$ and simplify using log laws ( $\log(ab)=\log a+\log b$ , $\log(a^p)=p\log a$ ).
Differentiate both sides: the left becomes $\dfrac{1}{y}\dfrac{dy}{dx}$ (chain rule), the right is now a sum.
Multiply back by $y$ : $\dfrac{dy}{dx} = y\cdot[\text{the differentiated right side}]$ .

The headline result is the variable-base, variable-exponent rule below — both the "treat the exponent as constant" term and the "treat the base as constant" term appear and ADD.

Derivative of f(x) raised to g(x)

\frac{d}{dx}\left[f(x)^{g(x)}\right] = f(x)^{g(x)}\left[g'(x)\,\log f(x) + g(x)\,\frac{f'(x)}{f(x)}\right]

$g'(x)\,\log f(x)$ the term from differentiating the exponent (treat base as constant)
g(x)\,f'(x)/f(x)the term from differentiating the base (treat exponent as constant)

Worked example

y = x^x

, find

\dfrac{dy}{dx}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q148Moderate

Example 1 · Differentiation · Logarithmic Differentiation

y=(\sin x)^{\tan x}

, then

\frac{dy}{dx}

is equal to

Both terms appear — never use just one

For

f^g

, the "power rule" alone (

g f^{g-1}f'

) and the "exponential rule" alone (

f^g\log f\cdot g'

) are each HALF the answer. Logarithmic differentiation produces BOTH terms and adds them. Using only one is the single most common error here.

A variable in the exponent kills the power rule

\dfrac{d}{dx}(x^x)

is NOT

x\cdot x^{x-1}

. The power rule

\dfrac{d}{dx}x^n = nx^{n-1}

requires

n

to be CONSTANT. When the exponent itself depends on

x

, take logs.

cos⁻¹(sin θ) collapses before you differentiate

In mixed stems like

\cos^{-1}(\sin\theta) + x^x

, use

\cos^{-1}(\sin\theta) = \tfrac{\pi}{2} - \theta

(for

\theta\in[0,\tfrac{\pi}{2}]

) to flatten the inverse-trig piece to a simple

-\theta'

; only the

x^x

part needs log differentiation. Mixing the two methods up wastes time.

Concept 2 of 5: Products, Quotients and Powers via Logs

When

y

is a tangle of products, quotients, and fractional powers, taking logs once flattens the whole thing into a SUM of simple

\log

terms. Each term differentiates to

\dfrac{(\text{inner})'}{\text{inner}}

, and you read off

dy/dx

almost by inspection.

Definition

After $\log y = \log(\text{everything})$ , use the three log laws to split:

Product $\to$ sum: $\log(ab) = \log a + \log b$ .
Quotient $\to$ difference: $\log(a/b) = \log a - \log b$ .
Power $\to$ coefficient: $\log(a^p) = p\log a$ (so a fractional exponent $4/3$ becomes a coefficient $4/3$ ).

Each resulting $\log u(x)$ differentiates to $\dfrac{u'(x)}{u(x)}$ ; then multiply the whole sum by $y$ . If the function is already wrapped in a log ( $y = \log(\cdots)$ ), you do NOT have a hidden $1/y$ — just expand the inside with log laws and differentiate the sum directly.

Log of a power-product

\log\!\left(\frac{a^{p}\,b^{q}}{c^{r}}\right) = p\log a + q\log b - r\log c

Worked example

y = \dfrac{x^2\sqrt{x-1}}{(x+2)^3}

, find

\dfrac{dy}{dx}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q110Moderate

Example 2 · Differentiation · Logarithmic Differentiation

y=\log_e\!\left(\dfrac{x^{5}\cdot(3x-4)^{4/3}}{(x+5)^{4/3}}\right)

, then

\dfrac{dy}{dx}

is equal to

If y is already a log, there is no 1/y

For

y = \log(\text{expression})

, expand the inside with log laws and differentiate the SUM directly. The

\dfrac{1}{y}\dfrac{dy}{dx}

form is only for

\log y = \cdots

(i.e. you took the log yourself).

Simplify before you differentiate — $\log\sqrt{\frac{1+\sin x}{1-\sin x}}$

Some quotient-log stems collapse to a tidy single function.

\log\sqrt{\tfrac{1+\sin x}{1-\sin x}} = \log\tan\!\left(\tfrac{\pi}{4}+\tfrac{x}{2}\right)

, whose derivative is the clean

\sec x

. Charging in with the quotient rule on the raw fraction works but is far slower and error-prone.

A fractional exponent becomes a fractional COEFFICIENT

\log[(x+5)^{4/3}] = \tfrac{4}{3}\log(x+5)

, not

\tfrac{4}{3}(x+5)

and not

(x+5)^{4/3}\log

. Drop the exponent out front; do not leave it on the argument.

Drill 3 more on products, quotients and powers via logs

Concept 3 of 5: The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0

This is the chapter's signature trick. A long product like

[(x+1)(2x+1)\cdots(nx+1)]^p

looks frightening, but logs turn it into a sum and the magic happens AT

x=0

: every factor

kx+1

equals

1

there, so

y=1

and the messy denominators all collapse, leaving a simple sum like

\sum k

\sum k^2

Definition

For $y = [(x+1)(2x+1)(3x+1)\cdots(nx+1)]^{p}$ :

Take logs: $\log y = p\sum_{k=1}^{n}\log(kx+1)$ .
Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx} = p\sum_{k=1}^{n}\dfrac{k}{kx+1}$ .
Evaluate at $x=0$ : every $kx+1=1$ so $y=1$ and $\dfrac{dy}{dx} = p\sum_{k=1}^{n}k$ .

The leftover sum is a standard power-sum (carried in the formula box). If the $k$ -th factor is $k^2x+1$ instead, you get $\sum k^2$ . If the product runs $(1-x)(2-x)\cdots(n-x)$ and you evaluate at $x=1$ , only the vanishing factor's term survives — handle that by factoring it out, not by the sum.

Power sums (the leftover at x=0)

\sum_{k=1}^{n} k = \frac{n(n+1)}{2} \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

pthe outer power (e.g. $2$ , $4$ , $3/2$ , or $n$ ) — it just multiplies the sum
kthe coefficient of $x$ in the $k$ -th factor; squared factors give $\sum k^2$

Worked example

y = (1+2x)(1+5x)(1+7x)

, find

\dfrac{dy}{dx}

x=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q101Moderate

Example 3 · Differentiation · Logarithmic Differentiation

y = [(x+1)(2x+1)(3x+1)\cdots(nx+1)]^{2}

, then

\frac{dy}{dx}

x=0

Substitute x=0 only AFTER differentiating

If you plug

x=0

into

y

before differentiating, you get the constant

1

and derivative

0

— wrong. Differentiate the log-sum fully (keep

x

), THEN set

x=0

so the denominators collapse to

1

Squared factor $\Rightarrow$ $\sum k^2$ , not $\sum k$

If the

k

-th factor is

k^2x+1

(i.e.

1, 4x+1, 9x+1,\ldots

), differentiating

\log(k^2x+1)

gives

\dfrac{k^2}{k^2x+1}

, so the leftover sum is

\sum k^2 = \tfrac{n(n+1)(2n+1)}{6}

. Don't reflexively write

\sum k

Product like (1-x)(2-x)⋯(n-x) at x=1 — factor, don't sum

When a factor itself VANISHES at the evaluation point (here

(1-x)=0

x=1

), the

\dfrac{1}{y}y'

form blows up. Instead write

y=(1-x)\,g(x)

; then

y'(1) = -1\cdot g(1)

, where

g(1)

is the product of the remaining factors at

x=1

The outer power just multiplies the sum

A power

3/2

n

on the whole product becomes a coefficient

p

\log y = p\sum\log(\cdots)

. So the answer is always

p\sum k

(or

p\sum k^2

) — e.g. power

3/2

gives

\tfrac{3}{2}\cdot\tfrac{n(n+1)}{2} = \tfrac{3n(n+1)}{4}

Drill 6 more on the product chain [(x+1)(2x+1)⋯(nx+1)] evaluated at x=0

Concept 4 of 5: Change of Base and log-of-a-log Forms

A logarithm with a VARIABLE base — like

\log_{\sin x}(\tan x)

— can't be differentiated as is. Change it to natural logs first via

\log_a b = \dfrac{\log b}{\log a}

; now both top and bottom are ordinary logs and the quotient rule finishes the job.

Definition

Change of base: $\log_a b = \dfrac{\log b}{\log a}$ (any common base; use natural $\log$ ). This converts a variable-base logarithm into a QUOTIENT of two natural logs, after which differentiate by the quotient rule. Nested form: $\log_{x^2}(\log x) = \dfrac{\log(\log x)}{\log(x^2)} = \dfrac{\log(\log x)}{2\log x}$ ; differentiate this quotient, then evaluate. Many of these are asked at a clean point ( $x = \tfrac{\pi}{4}$ , $x = e$ ) where one of the two log terms vanishes, killing half the quotient-rule expression.

Change of base

\log_a b = \frac{\log b}{\log a}

$\log$ natural log (base $e$ ) throughout this chapter
athe base — when it depends on $x$ , this is why you must convert

Worked example

y = \log_{x}(e)

, find

\dfrac{dy}{dx}

for

x>0,\ x\neq 1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q102Hard

Example 4 · Differentiation · Logarithmic Differentiation

y = \log_{\sin x}(\tan x)

, then

\left.\frac{dy}{dx}\right|_{x=\pi/4}

has the value

Convert the variable base BEFORE differentiating

\dfrac{d}{dx}\log_{\sin x}(\tan x)

cannot be done with the

\dfrac{1}{u}

rule directly — the base

\sin x

is not constant. Always rewrite as

\dfrac{\log\tan x}{\log\sin x}

first, then use the quotient rule.

A vanishing log term kills half the quotient rule

At nice points (

x=\pi/4

gives

\log\tan x = 0

;

x=e

gives

\log\log x = 0

), the quotient-rule term multiplying that zero disappears. Spot the vanishing term FIRST — it saves most of the algebra. The bank answer for

\log_{\sin x}\tan x

\pi/4

-4\log 2

log of a log is NOT (log)²

\log(\log x)

is a composition (log applied to

\log x

), with derivative

\dfrac{1}{\log x}\cdot\dfrac{1}{x}

. It is not

(\log x)^2

, whose derivative would be

2\log x\cdot\tfrac{1}{x}

. Keep the two straight.

Drill 2 more on change of base and log-of-a-log forms

Concept 5 of 5: Square-Root Quotients with Inverse-Trig Arguments

A function like

\sqrt{\dfrac{1-\sin^{-1}x}{1+\sin^{-1}x}}

bundles a square root, a quotient, and an inverse-trig term. Logs flatten all three at once: the

\sqrt{\ }

becomes a

\tfrac{1}{2}

coefficient and the quotient becomes a difference, leaving a short sum to differentiate and evaluate.

Definition

For $y = \sqrt{\dfrac{1 - u}{1 + u}}$ where $u = u(x)$ (e.g. $u = \sin^{-1}x$ ):

Take logs and use the square root $\to$ $\tfrac{1}{2}$ : $\log y = \tfrac{1}{2}[\log(1-u) - \log(1+u)]$ .
Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx} = \tfrac{1}{2}\left[\dfrac{-u'}{1-u} - \dfrac{u'}{1+u}\right]$ .
Multiply back by $y$ and evaluate at the given point.

When asked at $x=0$ , note $\sin^{-1}0 = 0$ , so $u=0$ , $y=1$ , and $u' = \dfrac{1}{\sqrt{1-x^2}} = 1$ at $x=0$ — the whole expression collapses to a single clean number.

Log of a square-root quotient

\log\sqrt{\frac{1-u}{1+u}} = \frac{1}{2}\Big[\log(1-u) - \log(1+u)\Big]

uthe inner function, e.g. $\sin^{-1}x$ , with $u' = 1/\sqrt{1-x^2}$
1/2the coefficient produced by the outer square root

Worked example

y = \sqrt{\dfrac{1-x}{1+x}}

, find

\dfrac{dy}{dx}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q150Moderate

Example 5 · Differentiation · Logarithmic Differentiation

y = \sqrt{\frac{1-\sin^{-1}(x)}{1+\sin^{-1}(x)}}

, then

\frac{dy}{dx}

x=0

and

y=1

Don't forget the chain factor u' on the inverse-trig inner

When

u = \sin^{-1}x

, each

\log(1\pm u)

differentiates to

\dfrac{\pm u'}{1\pm u}

with

u' = \dfrac{1}{\sqrt{1-x^2}}

. Dropping the

u'

(treating

u

x

) is a frequent error; at

x=0

it happens to equal

1

, but you must include it in general.

Compute y at the point — usually y=1 at x=0

You multiply

\dfrac{1}{y}y'

y

to finish. At

x=0

the quotient under the root is

\tfrac{1}{1}=1

, so

y=1

and the multiply-back is trivial. Forgetting to put

y

back (leaving only

\tfrac{1}{y}y'

) gives the wrong magnitude when

y\neq 1

Watch which factor is on top — it sets the sign

\sqrt{\tfrac{1-\sin^{-1}x}{1+\sin^{-1}x}}

gives

-1

x=0

; flipping to

\sqrt{\tfrac{1+\sin^{-1}x}{1-\sin^{-1}x}}

gives

+1

. The numerator/denominator order flips every sign — read the stem carefully before reaching for a memorised answer.

Drill 1 more on square-root quotients with inverse-trig arguments

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 (5)

Logarithmic Differentiation — the Method
Derivative of f(x) raised to g(x)
$\frac{d}{dx}\left[f(x)^{g(x)}\right] = f(x)^{g(x)}\left[g'(x)\,\log f(x) + g(x)\,\frac{f'(x)}{f(x)}\right]$
Products, Quotients and Powers via Logs
Log of a power-product
$\log\!\left(\frac{a^{p}\,b^{q}}{c^{r}}\right) = p\log a + q\log b - r\log c$
The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0
Power sums (the leftover at x=0)
$\sum_{k=1}^{n} k = \frac{n(n+1)}{2} \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
Change of Base and log-of-a-log Forms
Change of base
$\log_a b = \frac{\log b}{\log a}$
Square-Root Quotients with Inverse-Trig Arguments
Log of a square-root quotient
$\log\sqrt{\frac{1-u}{1+u}} = \frac{1}{2}\Big[\log(1-u) - \log(1+u)\Big]$

Watch out for (16)

Both terms appear — never use just one
→ Logarithmic Differentiation — the Method
A variable in the exponent kills the power rule
→ Logarithmic Differentiation — the Method
cos⁻¹(sin θ) collapses before you differentiate
→ Logarithmic Differentiation — the Method
If y is already a log, there is no 1/y
→ Products, Quotients and Powers via Logs
Simplify before you differentiate — $\log\sqrt{\frac{1+\sin x}{1-\sin x}}$
→ Products, Quotients and Powers via Logs
A fractional exponent becomes a fractional COEFFICIENT
→ Products, Quotients and Powers via Logs
Substitute x=0 only AFTER differentiating
→ The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0
Squared factor $\Rightarrow$ $\sum k^2$ , not $\sum k$
→ The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0
Product like (1-x)(2-x)⋯(n-x) at x=1 — factor, don't sum
→ The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0
The outer power just multiplies the sum
→ The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0
Convert the variable base BEFORE differentiating
→ Change of Base and log-of-a-log Forms
A vanishing log term kills half the quotient rule
→ Change of Base and log-of-a-log Forms
log of a log is NOT (log)²
→ Change of Base and log-of-a-log Forms
Don't forget the chain factor u' on the inverse-trig inner
→ Square-Root Quotients with Inverse-Trig Arguments
Compute y at the point — usually y=1 at x=0
→ Square-Root Quotients with Inverse-Trig Arguments
Watch which factor is on top — it sets the sign
→ Square-Root Quotients with Inverse-Trig Arguments

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