Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.2 | Set 1

Last Updated : 30 Apr, 2021

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Question 1. Differentiate each of the following using first principles:

(i) 2/x

Solution:

Given that f(x) = 2/x

By using the formula

f'(x) = $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{2}{x+h}-\frac{2}{x}}{h}$

= $\lim_{h\to 0}\frac{2x-2x-2h}{hx(x+h)}$

= $\lim_{h\to 0}\frac{2(-h)}{hx(x+h)}$

= $\lim_{h\to 0}\frac{-2}{x(x+h)}$

= $\frac{-2}{x^2}$

(ii) 1/√x

Solution:

Given that f(x) = 1/√x

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{1}{\sqrt(x+h)}-\frac{1}{\sqrt{x}}}{h}$

= $\lim_{h\to 0}\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}.h}$

= $\lim_{h\to 0}\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}.h}\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}.h}$

= $\lim_{h\to 0}\frac{x-(x+h)}{\sqrt{x}\sqrt{x+h}.h(\sqrt{x}+\sqrt{x+h})}$

= $\lim_{h\to 0}\frac{-h}{\sqrt{x}\sqrt{x+h}.h(\sqrt{x}+\sqrt{x+h})}$

= $\lim_{h\to 0}\frac{-1}{x\sqrt{x+h}+\sqrt{x}(\sqrt{x+h})}$

= $\lim_{h\to 0}\frac{-1}{2x\sqrt{x}}$

= $\frac{-1}{2}x^{-\frac{3}{2}}$

(iii) 1/x³

Solution:

We have f(x) = 1/x³

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{1}{(x+h)^3}-\frac{1}{x^3}}{h}$

= $\lim_{h\to 0}\frac{x^3-(x^3+3x^2h+3xh^2+h^3)}{x^3h(x+h)^3}$

= $\lim_{h\to 0}\frac{x^3-x^3-3x^2-3xh-h^2}{x^3(x+h)^3}$

= $\lim_{h\to 0}\frac{-3x^2-3xh-h^2}{x^3(x+h)^3}$

= $\frac{-3x^2}{x^6}$

= $\frac{-3}{x^4}$

(iv) (x²+ 1)/x

Solution:

Given that f(x) = (x²+ 1)/x

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{(x+h)^2+1}{(x+h)}-\frac{x^2+1}{x}}{h}$

= $\lim_{h\to 0}\frac{x[x^2+h^2+2xh+1]-(x^2+1)(x+h)}{hx(x+h)}$

= $\lim_{h\to 0}\frac{x^3+xh^2+2x^2h+x-x^3-x-x^2h-h}{hx(x+h)}$

= $\lim_{h\to 0}\frac{xh+2x^2-x^2-1}{x(x+h)}$

= $\frac{x^2-1}{x^2}$

= $1-\frac{1}{x^2}$

(v) (x²– 1)/x

Solution:

Given that f(x) = (x²– 1)/x

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{(x+h)^2-1}{(x+h)}-\frac{x^2-1}{x}}{h}$

= $\lim_{h\to 0}\frac{x[x^2+h^2+2xh-1]-(x^2-1)(x+h)}{hx(x+h)}$

= $\lim_{h\to 0}\frac{xh+2x^2-x^2+1}{x(x+h)}$

= $\frac{x^2+1}{x^2}$

= $1+\frac{1}{x^2}$

(vi) (x + 1)/(x + 2)

Solution:

Given that f(x) = (x + 1)/(x + 2)

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{(x+h)+1}{(x+h)+2}-\frac{x+1}{x+2}}{h}$

= $\lim_{h\to 0}\frac{(x^2+2x+xh+2h+2+x)-(x^2+xh+2x+x+h+2)}{h(x+2)(x+h+2)}$

= $\lim_{h\to 0}\frac{h}{h(x+2)(x+h+2)}$

= 1/(x + 2)²

(vii) (x + 2)/(3x + 5)

Solution:

Given that f(x) = (x + 2)/(3x + 5)

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{(x+h+2)}{3(x+h)+5}-\frac{x+2}{3x+5}}{h}$

= $\lim_{h\to 0}\frac{(3x+5)(x+h+2)-(x+2)(3x+3h+5)}{h(3x+5)(3x+3h+5)}$

= $\lim_{h\to 0}\frac{(3x^2+5x+3xh+5h+6x+10)-(3x^2+3xh+5x+6x+6h+10)}{h(3x+5)(3x+3h+5)}$

= $\lim_{h\to 0}\frac{-h}{h(3x+5)(3x+3h+5)}$

= $\lim_{h\to 0}\frac{-1}{(3x+5)(3x+3h+5)}$

= $\frac{-1}{(3x+5)^2}$

(viii) kxⁿ

Solution:

Given that f(x) = kxⁿ

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{k(x+h)^n-kx^n}{h}$

= $k\lim_{h\to 0}\frac{(x^n+nx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h^2+...)-x^n}{h}$

= $k\lim_{h\to 0}nx^{n-1}+\frac{n(n-1)}{2!}x^{n-2}h+\frac{n(n-1(n-2)}{3!}x^{n-3}h^2......$

= k nx^n-1+ 0 + 0 …

= k nx^n-1

(ix) 1/√(3 – x)

Solution:

Given that f(x) = 1/√(3-x)

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{1}{\sqrt{3-(x+h)}}-\frac{1}{\sqrt{3-x}}}{h}$

= $\lim_{h\to 0}\frac{\sqrt{3-x}-\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}h}$

= $\lim_{h\to 0}\frac{\sqrt{3-x}-\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}h}\frac{\sqrt{3-x}+\sqrt{3-(x+h)}}{\sqrt{3-x}+\sqrt{3-(x+h)}h}$

= $\lim_{h\to 0}\frac{(3-x)-(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}h\sqrt{3-x}+\sqrt{3-(x+h)}}$

= $\lim_{h\to 0}\frac{h}{\sqrt{3-x}\sqrt{3-(x+h)}h(\sqrt{3-x}+\sqrt{3-(x+h))}}$

= $\frac{1}{(3-x)2\sqrt{3-x}}$

= $\frac{1}{2(3-x)^{3/2}}$

(x) x²+ x + 3

Solution:

Given that f(x) = x²+ x + 3

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{[(x+h)^2+(x+h)+3]-x^2+x+3}{h}$

= $\lim_{h\to 0}\frac{x^2+h^2+2hx+x+h+3-x^2-x-3}{h}$

= $\lim_{h\to 0}\frac{2xh+h^2+h}{h}$

= $\lim_{h\to 0}\frac{h(2x+h+1)}{h}$

= 2x + 0 + 1

= 2x + 1

(xi) (x + 2)³

Solution:

Given that f(x) = (x + 2)³

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{(x+h+2)^3-(x+2)^3}{h}$

= $\lim_{h\to 0}\frac{[(x+2)+h]^3-(x+2)^3}{h}$

= $\lim_{h\to 0}\frac{(x+2)^3+h^3+3h(x+2)^2+3(x+2)h^2-(x+2)^3}{h}$

= $\lim_{h\to 0}3(x+2)^2+3(x+2)h+h^2$

= 3(x + 2)²

(xii) x³+ 4x² + 3x + 2

Solution:

Given that f(x) = x³+ 4x² + 3x + 2

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{(x+h)^3+4(x+h)^2+3(x+h)+2-(x^3+4x^2+3x+2)}{h}$

= $\lim_{h\to 0}\frac{x^3+h^3+3x^2h+3h^2x+4x^2+4h^2+8hx+3x+3h+2-x^3-4x^2-3x-2}{h}$

= $\lim_{h\to 0}\frac{3x^2h+3h^2x+h^3+4h^2+8hx+3h}{h}$

= $\lim_{h\to 0}3x^2+3hx+h^2+4h+8x+3$

= 3x²+ 8x + 3

(xiii) (x²+ 1)(x – 5)

Solution:

Given that f(x) = (x²+1)(x-5)

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{[(x+h)^3+(x+h)-5(x+h^2)-5]-(x^3-5x^2+x-5)}{h}$

= $\lim_{h\to 0}\frac{(x^3+h^3+3x^2h+3h^2x+x+h-5x^2-5h^3-10xh-5)-(x^3-5x^2+x-5)}{h}$

= $\lim_{h\to 0}\frac{(3x^2h+3h^2x+h^3+h-5h^2-10xh)}{h}$

= $\lim_{h\to 0}3x^2+3hx+h^2+1-5h-10x$

= 3x²– 10x + 1

(xiv) √(2x²+ 1)

Solution:

Given that f(x) = √(2x²+ 1)

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\sqrt{2(x+h)^2+1}-\sqrt{2x^2+1}}{h}$

On multiplying numerator and denominator by ${\sqrt{2(x+h)^2+1}+\sqrt{2x^2+1}}$

We get

= $\lim_{h\to 0}\frac{[2(x+h)^2+1-(2x^2+1)]}{h(\sqrt{2(x+h)^2+1}+\sqrt{2x^2+1})}$

= $\lim_{h\to 0}\frac{2x^2+2h^2+4xh+1-2x^2-1}{h(\sqrt{2(x+h)^2+1}+\sqrt{2x^2+1})}$

= $\lim_{h\to 0}\frac{4xh+2h^2}{h\sqrt{2(x+h)^2+1}+\sqrt{2x^2+1}}$

= $\frac{4x}{2\sqrt{2x^2+1}}$

= $\frac{2x}{\sqrt{2x^2+1}}$

(xv) (2x + 3)/(x – 2)

Solution:

Given that f(x) = (2x + 3)/(x – 2)

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{\frac{2x+2h+3}{x+h-2}-\frac{2x+3}{2-x}}{h}$

= $\lim_{h\to 0}\frac{2x^2+2hx+3x-4x-4h-6-2x^2-2hx+4x-3x-3h+6}{h(x+h-2)(x-2)}$

= $\lim_{h\to 0}\frac{-7}{(x+h-2)(x-2)}$

= $\frac{-7}{(x-2)^2}$

Question 2. Differentiate each of the following using first principles:

(i) e^-x

Solution:

Given that f(x) = e^-x

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $f'(x)=\lim_{h\to 0}\frac{e^{-(x+h)-e^{-x}}}{h}$

= $f'(x)=\lim_{h\to 0}\frac{e^{-x}(e^{-h}-1)}{h}$

= $f'(x)=\lim_{h\to 0}\frac{-e^{-x}(e^{-h}-1)}{-h}$

= -e^-x

(ii) e^3x

Solution:

Given that f(x) = e^3x

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{e^{3(x+h)}-e^{3x}}{h}$

= $\lim_{h\to 0}\frac{e^{3x}e^{3h}-e^{3x}}{h}$

= $\lim_{h\to 0}\frac{e^{3x}(e^{3h-1})}{h}$

Multiplying numerator and denominator by 3.

= $\lim_{h\to 0}e^{3x}\frac{(e^{3h-1})}{3h}$

Here, $\lim_{h\to 0}\frac{e^{3h-1}}{3h}=1$

= 3e^3x

(iii) e^ax+b

Solution:

Given that f(x) = e^ax+b

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{e^{x(x+h)+b}-e^{ax+b}}{h}$

= $\lim_{h\to 0}\frac{e^{ax}e^{ah}e^b-e^{ax}e^b}{h}$

= $\lim_{h\to 0}\frac{e^be^{ax}(e^{ah}-1)}{h}$

= $lim_{h\to 0}e^{(ax+b)}\frac{a(e^{ah}-1)}{a.h}$

On multiplying numerator and denominator by a

Since $\lim_{h\to 0}\frac{(e^{ah}-1)}{ah}=1$

= ae^ax+b

(iv) xe^x

Solution:

Given that f(x) = xe^x

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{(x+h)e^{x+h}-xe^x}{h}$

= $\lim_{h\to 0}\frac{xe^xe^h+he^xe^h-xe^x}{h}$

= $\lim_{h\to 0}xe^x(\frac{e^h-1}{h})+\frac{he^{x+h}}{h}$

= xe^x+ e^x

= e^x(x + 1)

(v) x²e^x

Solution:

Given that f(x) = x²e^x

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{x^2e^xe^h+h^2e^xe^h+2xhe^h-x^2e^x}{h}$

= $\lim_{h\to 0}x^2e^x\frac{(e^h-1)}{h}+e^xe^h\frac{(h^2+2xh)}{h}$

= x²e^x + e^x(0 + 2x)

= x²e^x + 2xe^x

= e^x(x² + 2x)

(vi) $e^{(x^2+1)}$

Given that f(x) = $e^{(x^2+1)}$

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{e^{(x+h)^2}+1-e^{x^2+1}}{h}$

= $\lim_{h\to 0}\frac{e^{x^2+h^2+2xh+1}-e^{x^2+1}}{h}$

= $\lim_{h\to 0}\frac{e^{x^2+1}(e^{2xh}e^{h^2}-1)}{h}$

= $\lim_{h\to 0}\frac{e^{x^2+1}(e^{2xh+h^2}-1)}{2xh+h^2}×\frac{2xh+h^2}{h}$

= $\lim_{h\to 0}e^{x^2+1}.1×2x+h$

= $2xe^{x^2+1}$

(vii) e^√(2x)

Solution:

Given that f(x) = $e^{\sqrt{2x}}$

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}$

= $\lim_{h\to 0}\frac{e^{\sqrt{2x}}(e^{\sqrt{2(x+h)}-\sqrt{2x}}-1)}{h}$

= $\lim_{h\to 0}e^{\sqrt{2x}}\frac{(e^{\sqrt{2(x+h)}-\sqrt{2x}}-1)}{\sqrt{2(x+h)}-\sqrt{2x}}\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h}$

On multiplying numerator and denominator by $\sqrt{2(x+h)}-\sqrt{2x}$

we get

= $\lim_{h\to 0}e^{\sqrt{2x}}\frac{{\sqrt{2(x+h)}-\sqrt{2x}}}{h}$

Again multiplying numerator and denominator by $\sqrt{2(x+h)}+\sqrt{2x}$

we get

= $\lim_{h\to 0}e^{\sqrt{2x}}\frac{{\sqrt{2(x+h)}-\sqrt{2x}}}{h}×\frac{\sqrt{2(x+h)}+\sqrt{2x}}{\sqrt{2(x+h)}+\sqrt{2x}}$

= $e^{\sqrt{2x}}×\frac{1}{2\sqrt{2x}}$

(viii) e^{√(ax + b)}

Solution:

Given that f(x) = e^√(ax+b)

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{e^{\sqrt{a(x+h)+b}}-e^{\sqrt{ax+b}}}{h}$

= $\lim_{h\to 0}e^{\sqrt{a(x+h)+b}}\frac{(e^{\sqrt{a(x+h)+b}-{\sqrt{ax+b}}}-1)}{h}$

= $\lim_{h\to 0}e^{\sqrt{ax+b}}\frac{e^{\sqrt{a(x+h)+b}-{\sqrt{ax+b}}}-1}{\sqrt{a(x+h)+b}-\sqrt{ax+b}}\frac{\sqrt{a(x+h)+b}-\sqrt{ax+b}}{h}$

On multiplying numerator and denominator by $\sqrt{a(x+h)+b}-\sqrt{ax+b}$

we get

= $\lim_{h\to 0}e^{\sqrt{ax+b}}×1×\frac{{\sqrt{a(x+h)+b}-{\sqrt{ax+b}}}}{\sqrt{a(x+h)+b}+\sqrt{ax+b}}×\frac{\sqrt{a(x+h)+b}+\sqrt{ax+b}}{h}$

Again multiplying numerator and denominator by $\sqrt{a(x+h)+b}+\sqrt{ax+b}$

we get

= $\lim_{h\to 0}e^{\sqrt{ax+b}}×\frac{a(x+h)+b-(ax+b)}{h}×\frac{1}{(\sqrt{a(x+h)+b}+\sqrt{ax+b})}$

= $\frac{e^{\sqrt{ax+b}}×a}{2\sqrt{ax+b}}$

= $\frac{ae^{\sqrt{ax+b}}}{2\sqrt{ax+b}}$

(ix) a^√x

Solution:

Given that f(x) = a^√x = e^√xloga

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{e^{\sqrt{x+h}loga}-e^{\sqrt{x}loga}}{h}$

= $\lim_{h\to 0}e^{\sqrt{x}loga}\frac{e^{\sqrt{x+h}loga-\sqrt{x}loga}-1}{h}$

= $\lim_{h\to 0}e^{\sqrt{x}loga}\frac{e^{(\sqrt{x+h}-\sqrt{x})loga}-1}{h}$

On multiplying numerator and denominator by $(\sqrt{x+h}-\sqrt{x})loga$

we get

f”(x) = $\lim_{h\to 0}e^{\sqrt{x}loga}\frac{e^{(\sqrt{x+h}-\sqrt{x})loga}-1}{h(\sqrt{x+h}-\sqrt{x})loga}(\sqrt{x+h}-\sqrt{x})loga$

= $e^{\sqrt{x}loga}\lim_{h\to 0}\frac{e^{(\sqrt{x+h}-\sqrt{x})loga}-1}{(\sqrt{x+h}-\sqrt{x})loga}\lim_{h\to 0}loga\frac{(\sqrt{x+h}-\sqrt{x})}{h}$

= $e^{\sqrt{x}loga}\lim_{h\to 0}loga\frac{(\sqrt{x+h}-\sqrt{x})}{h}$

On multiplying numerator and denominator by $(\sqrt{x+h}+\sqrt{x})$

we get

f'(x) = $e^{\sqrt{x}loga}\lim_{h\to 0}loga\frac{(\sqrt{x+h}-\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}(\sqrt{x+h}+\sqrt{x})$

= $e^{\sqrt{x}loga}\lim_{h\to 0}loga\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$

= $e^{\sqrt{x}loga}\frac{loga}{2\sqrt{x}}$

= $\frac{a^{\sqrt{x}}}{2\sqrt{x}}$ log_ea

(x) $3^{x^2}$

Solution:

Given that f(x) = $3^{x^2}=e^{x^2log3}$

By using the formula

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

We get

= $\lim_{h\to 0}\frac{e^{(x+h)^2log3}-e^{x^2log3}}{h}$

= $\lim_{h\to 0}e^{x^2log3}\frac{[e({(x+h)^2-x^2)^{log3}}-1]}{h}$

= $\lim_{h\to 0}e^{x^2log3}\frac{[e{(x+h)^2-x^2]^{log3}}-1}{(x+h)^2-x^2}×\frac{(x+h)^2-x^2}{h}$

= $\lim_{h\to 0}e^{x^2log3}\frac{(x+h+x)(x+h-x)}{h}$

= $e^{x^2log3} 2x$

= $2xe^{x^2log3}$

= $2x3^{x^2log3}$