Exercise 5.5 Class 12 Maths Continuity and Differentiability

NCERT Solutions Class 12 Maths Chapter 5 Ex 5.5 Detailed | LearnCBSEHub

💡 When to use Logarithms?

1. Variable to variable power: $y = x^x, (\sin x)^{\cos x}$, etc.

2. Complex Products/Quotients: $y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}}$

Rule: Take $\log$ on both sides: $\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$

Question 01

Differentiate $y = \cos x \cdot \cos 2x \cdot \cos 3x$

Step 1: Take Logarithm

$$\log y = \log(\cos x) + \log(\cos 2x) + \log(\cos 3x)$$

Step 2: Differentiate

$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{\cos x}(-\sin x) + \frac{1}{\cos 2x}(-\sin 2x) \cdot 2 + \frac{1}{\cos 3x}(-\sin 3x) \cdot 3$$ $$\frac{1}{y}\frac{dy}{dx} = -(\tan x + 2\tan 2x + 3\tan 3x)$$

Answer: $-y(\tan x + 2\tan 2x + 3\tan 3x)$

Question 02

Differentiate $y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Step 1: Logarithm Properties

Using $\log \sqrt{A} = \frac{1}{2} \log A$ and $\log(A/B) = \log A – \log B$.

$$\log y = \frac{1}{2}[\log(x-1) + \log(x-2) – \log(x-3) – \log(x-4) – \log(x-5)]$$

Step 2: Differentiate

$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\left[\frac{1}{x-1} + \frac{1}{x-2} – \frac{1}{x-3} – \frac{1}{x-4} – \frac{1}{x-5}\right]$$

Answer: $\frac{y}{2}\left[\frac{1}{x-1} + \frac{1}{x-2} – \frac{1}{x-3} – \frac{1}{x-4} – \frac{1}{x-5}\right]$

Question 03

Differentiate $y = (\log x)^{\cos x}$

Step 1: Take Log

$\log y = \cos x \cdot \log(\log x)$

Step 2: Product Rule

$$\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot (-\sin x)$$ $$\frac{dy}{dx} = y \left[\frac{\cos x}{x \log x} – \sin x \log(\log x)\right]$$

Answer: $(\log x)^{\cos x} \left[\frac{\cos x}{x \log x} – \sin x \log(\log x)\right]$

Question 04

Differentiate $y = x^x – 2^{\sin x}$

Step 1: Separate terms

Let $y = u – v$, where $u = x^x$ and $v = 2^{\sin x}$.

For u: $\log u = x \log x \implies \frac{1}{u}\frac{du}{dx} = 1 + \log x \implies u’ = x^x(1+\log x)$

For v: $v = 2^{\sin x} \implies v’ = 2^{\sin x} \cdot \log 2 \cdot \cos x$

Answer: $x^x(1+\log x) – 2^{\sin x} \cos x \log 2$

Question 05

Differentiate $y = (x+3)^2 (x+4)^3 (x+5)^4$

Step 1: Logarithm

$$\log y = 2\log(x+3) + 3\log(x+4) + 4\log(x+5)$$

Step 2: Differentiate

$$\frac{1}{y}\frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}$$

Answer: $y\left[\frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}\right]$

Question 06

Differentiate $y = \left(x + \frac{1}{x}\right)^x + x^{\left(1 + \frac{1}{x}\right)}$

Step 1: Split into u and v

Let $u = (x + 1/x)^x$. Taking log: $\log u = x \log(x + 1/x)$
$u’ = (x + 1/x)^x \left[\log(x + 1/x) + \frac{x^2-1}{x^2+1}\right]$

Let $v = x^{(1+1/x)}$. Taking log: $\log v = (1+1/x)\log x$
$v’ = x^{(1+1/x)} \left[\frac{x+1}{x^2} – \frac{\log x}{x^2}\right]$

Answer: Sum of derivatives of u and v.

Question 07

Differentiate $y = (\log x)^x + x^{\log x}$

Step 1: Calculate Terms

$\frac{d}{dx}[(\log x)^x] = (\log x)^x \left[\frac{1}{\log x} + \log(\log x)\right]$

$\frac{d}{dx}[x^{\log x}] = \frac{d}{dx}[e^{(\log x)^2}] = x^{\log x} \left[\frac{2\log x}{x}\right]$

Answer: Sum of the two terms.

Question 08

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

Step 1: Calculate

$\frac{d}{dx}[(\sin x)^x] = (\sin x)^x [x \cot x + \log \sin x]$

$\frac{d}{dx}[\sin^{-1}\sqrt{x}] = \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x-x^2}}$

Answer: $(\sin x)^x [x \cot x + \log \sin x] + \frac{1}{2\sqrt{x-x^2}}$

Question 09

Differentiate $y = x^{\sin x} + (\sin x)^{\cos x}$

Step 1: Use Log Differentiation

$u = x^{\sin x} \implies u’ = x^{\sin x}[\frac{\sin x}{x} + \cos x \log x]$
$v = (\sin x)^{\cos x} \implies v’ = (\sin x)^{\cos x}[\cos x \cot x – \sin x \log \sin x]$

Answer: $u’ + v’$

Question 10

Differentiate $y = x^{x \cos x} + \frac{x^2+1}{x^2-1}$

Step 1: Differentiate Terms

$u = x^{x \cos x} \implies \log u = x \cos x \log x$.
$u’ = x^{x \cos x} [\cos x \log x + \cos x – x \sin x \log x]$

$v = \frac{x^2+1}{x^2-1} \implies v’ = \frac{-4x}{(x^2-1)^2}$ (Quotient Rule)

Answer: $u’ + v’$

Question 11

Differentiate $y = (x \cos x)^x + (x \sin x)^{1/x}$

Step 1: Log Differentiation

$u = (x \cos x)^x \implies \log u = x[\log x + \log \cos x]$
$u’ = (x \cos x)^x [1 + \log(x \cos x) – x \tan x]$

$v = (x \sin x)^{1/x} \implies \log v = \frac{1}{x}[\log x + \log \sin x]$
$v’ = (x \sin x)^{1/x} \left[\frac{x \cot x + 1 – \log(x \sin x)}{x^2}\right]$

Answer: Sum of $u’$ and $v’$

Questions 12 – 15 • Implicit Differentiation

Find dy/dx for implicit relations involving powers.

Q12: $x^y + y^x = 1$
Let $u=x^y, v=y^x$. Then $u+v=1 \implies u’ + v’ = 0$.
$u’ = x^y(y/x + \log x \cdot y’)$ and $v’ = y^x(x/y \cdot y’ + \log y)$.
Solve for $y’$: $y’ = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$

Q13: $y^x = x^y$
Take log: $x \log y = y \log x$. Differentiate implicitly.
$1 \cdot \log y + x/y \cdot y’ = y’ \log x + y/x$.
$y’ = \frac{y(y – x \log y)}{x(x – y \log x)}$

Q14: $(\cos x)^y = (\cos y)^x$
Take log: $y \log \cos x = x \log \cos y$. Differentiate.
$y’ \log \cos x – y \tan x = \log \cos y – x \tan y \cdot y’$.
$y’ = \frac{y \tan x + \log \cos y}{x \tan y + \log \cos x}$

Q15: $xy = e^{x-y}$
Take log: $\log x + \log y = x – y$.
$1/x + 1/y \cdot y’ = 1 – y’$.
$y'(1/y + 1) = 1 – 1/x \implies y’ = \frac{y(x-1)}{x(y+1)}$

Question 16

If $f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)$, find $f'(1)$.

Step 1: Logarithm

$$\log f(x) = \log(1+x) + \log(1+x^2) + \dots$$ $$\frac{f'(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$$

Step 2: Value at x=1

$f(1) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$. Bracket term = $1/2 + 1 + 2 + 4 = 7.5 = 15/2$.

Answer: $f'(1) = 16 \times \frac{15}{2} = 120$.

Question 17

Differentiate $(x^2-5x+8)(x^3+7x+9)$ in three ways.

1. Product Rule: Differentiate as $uv$.
2. Expand Polynomial: Multiply to get a degree 5 polynomial, then differentiate.
3. Log Differentiation: Take log and differentiate.

Conclusion: All three methods yield the same result.

Question 18

Prove $\frac{d}{dx}(uvw) = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}$

Proof

Let $y = uvw$. Take $\log y = \log u + \log v + \log w$.

$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{u}\frac{du}{dx} + \frac{1}{v}\frac{dv}{dx} + \frac{1}{w}\frac{dw}{dx}$$ $$\frac{dy}{dx} = uvw \left[\frac{u’}{u} + \frac{v’}{v} + \frac{w’}{w}\right]$$ $$\frac{dy}{dx} = u’vw + uv’w + uvw’$$

Hence Proved.