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

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) x² e^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}

Summary

Exercise 30.2 Set 1 covers differentiation of various functions, including polynomials, trigonometric functions, exponential functions, and logarithmic functions. Students learn to apply differentiation rules and formulas to find derivatives. Understanding derivatives is crucial for calculus and its applications. Differentiation helps analyze functions and model real-world phenomena. Practice questions reinforce learning and application. Derivatives measure rates of change, essential for optimization and physics.