Class 11 RD Sharma Solutions - Chapter 30 Derivatives - Exercise 30.2 | Set 2

Derivatives are a fundamental concept in calculus that measures how a function changes as its input changes. They represent the rate of change or slope of the function at any given point. Understanding derivatives is crucial for solving problems related to motion, optimization, and modeling real-world phenomena. In Chapter 30 of RD Sharma's Class 11 textbook, students delve into the basics of derivatives including the rules and applications which are essential for mastering calculus.

Derivatives

The Derivatives provide a way to quantify the rate at which one quantity changes with respect to another. For a function f(x) the derivative 𝑓(x) represents the instantaneous rate of the change of f(x) concerning x. This concept is foundational in mathematics and is widely used in physics, engineering, economics, and various other fields. The Derivatives help in understanding the behavior of functions finding tangents to curves and solving the problems related to rates of change.

Question 3. Differentiate each of the following using first principles:

(i) x sinx

Solution:

Given that f(x) = xsinx

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)sin(x+h)-xsinx}{h}

= \lim_{h\to 0}\frac{x(sin(x+h)-sinx)}{h}+sin(x+h)                

Using the formula 

sinc - sind = 2cos((c + d)/2)sin((c - d)/2)

We get

= \lim_{h\to 0}\frac{x×2cos(x+h/2)sin(h/2)}{h}+sin(x+h)           

As we know that \lim_{θ\to 0}\frac{sinθ}{θ}=1

So, 

= 2x × cosx × 1/2 + sinx

= x × cosx + sinx

= sinx + xcosx

(ii) xcosx

Solution:   

Given that f(x) = xcosx

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)cos(x+h)-xcosx}{h}

= \lim_{h\to 0}\frac{xcos(x+h)hcos(x+h)-xcosx}{h}

= \lim_{h\to 0}\frac{x(cos(x+h)-cosx)}{h}+cos(x+h)

= \lim_{h\to 0}x.2sin(x-x+h/2)sin(x+h/2)+cos(x+h) [cosA-cosB=2sin(\frac{B-A}{2})sin(\frac{B+A}{2})]

= \lim_{h\to 0}\frac{2x.sin(-h/2)sin(x+h/2)}{h}+cos(x+h)

= -xsinx + cosx

(iii) sin(2x - 3)

Solution:

Given that f(x) = sin(2x - 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{sin[2(x+h)-3]-sin(2x-3)}{h}

= \lim_{h\to 0}\frac{2cos\frac{(2x+2h-3+2x-3)}{2}×sin\frac{(2x+2h-3-2-x+3)}{2}}{h} 

Using the formula

sinC - sinD = 2cos{C+D}/2sin{C-D}/2

= \lim_{h\to 0}2cos(2x-3+h).\frac{sinh}{h}               

As we know that, \lim_{θ\to 0}\frac{sinθ}{θ}=1 so,

= 2cos(2x - 3)

(iv) √sin2x

Solution:

Given that f(x) = √sin2x

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{sin2(x+h)}-\sqrt{sin2x}}{h}

On multiplying numerator and denominator by (\sqrt{sin2(x+h)}+\sqrt{sin2x})

we get

= \lim_{h\to 0}\frac{\sqrt{sin2(x+h)}-\sqrt{sin2x}}{h}×\frac{\sqrt{sin2(x+h)}+\sqrt{sin2x}}{\sqrt{sin2(x+h)}+\sqrt{sin2x}}

= \lim_{h\to 0}\frac{sin(2x+2h)-sin2x}{h(\sqrt{sin(2x+2h)}+\sqrt{sin2x})}

= \lim_{h\to 0}\frac{2cos(2x+2h)×sinh}{h}×\frac{1}{(\sqrt{sin(2x+2h)}+\sqrt{sin2x})}

= \frac{2cos2x}{2\sqrt{sin2x}}

= \frac{cos2x}{\sqrt{sin2x}}

(v) sinx/x

Solution:

Given that f{x} = sinx/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{sin(x+h)}{x+h}-\frac{sinx}{x})}{h}

= \lim_{h\to 0}\frac{xsin(x+h)-(x+h)sinx}{xh(x+h)}

= \lim_{h\to 0}\frac{x(sinx.cosh+cosx.sinh)-x.sinx-h.sinx}{xh(x+h)}

= \lim_{h\to 0}\frac{x.sinx(cosh-1)}{xh(x+h)}+\frac{x.cosx.sinh}{xh(x+h)}-\frac{hsinx}{(x+h)xh}

= \frac{-xsinx}{x(x+h)}×\frac{2sin^2h/2}{h^2/4}×h/4+\frac{xcosx}{x^2}-\frac{sinx}{x^2}

h ⇢ 0 ⇒ h/2 ⇢ 0 and \lim_{θ\to 0}\frac{sinθ}{θ}=1

= 0+\frac{xcosx-sinx}{x^2}

=\frac{xcosx-sinx}{x^2}

(vi) cosx/x

Solution:

Given that f(x) = cosx/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{cos(x+h)}{x+h}-\frac{cosx}{x}}{h}

= \lim_{h\to 0}\frac{x.cos(x+h)-(x+h)cosx}{h(x+h)}

= \lim_{h\to 0}\frac{x[cosx.cosh-sinx.sinh]-x.cos-h.cosx}{h(x+h)}

= \lim_{h\to 0}\frac{xcosx(cosh-1)}{h(x+h)}-\frac{x.cosx.sinh}{(x+h)h}-\frac{h.cosx}{(x+h)h}

= \lim_{h\to 0}\frac{-xcosx.2sin^2(h/2)}{(x+h)\frac{h^2}{4}}\frac{h^2}{4}-\frac{xsinx}{x(x+h)}-\frac{cosx}{x(x+h)}

= 0-\frac{xsinx}{x^2}-\frac{cosx}{x^2}

= -\frac{xsinx-cosx}{x^2}

(vii) x²sinx

Solution:

Given that f(x) = x²sinx

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)^2sin(x+h)-x^2sinx}{h}

= \lim_{h\to 0}\frac{(x^2+h^2+2hx)(sinx.cosh+cosx.sinh)-x^2sinx}{h}

= \lim_{h\to 0}\frac{x^2sinx(cosh-1)}{h}+\frac{h(h+2x)cosh.sinx}{h}-(x+h)^2cosx\frac{sinh}{h}

= \lim_{h\to 0}-x^2sinx\frac{2sin^2(h/2)}{(h/2)^2}h^2/4+(h+2x)sinx.cosh+(x+h)^2cosx

= 0 + [2xsinx + x²cosx]

= 2xsinx + x²cosx

(viii) \sqrt{sin(3x+1)}

Solution:

Given that f(x) = \sqrt{sin(3x+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{sin3(x+h)+1}-\sqrt{sin(3x+1)}}{h}

= \lim_{h\to 0}\frac{\sqrt{sin3(x+h)+1}-\sqrt{sin(3x+1)}}{h}\frac{\sqrt{sin3(x+h)+1}+\sqrt{sin(3x+1)}}{\sqrt{sin3(x+h)+1}+\sqrt{sin(3x+1)}}

= \lim_{h\to 0}\frac{sin(3x+3h+1)-sin(3x+1)}{h(\sqrt{sin3(x+h)+1}+\sqrt{sin(3x+1)})}

= \lim_{h\to 0}2cos(3x+1+3h/2)×\frac{sin(3h/2)}{3h/2}×3/2×\frac{1}{\sqrt{sin3(x+h)+1}+\sqrt{sin(3x+1)}}

= \frac{3cos(3x+1)}{2\sqrt{sin(3x+1)}}

(ix) sinx + cosx

Solution:

Given that f(x) = sinx + cosx

By using the formula

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

We get

= \lim_{h\to 0}\frac{[sin(x+h)+cos(x+h)]-sinx+cosx}{h}

= \lim_{h\to 0}\frac{[sin(x+h)+cos(x+h)-sinx-cosx]}{h}

= \lim_{h\to 0}\frac{[sin(x+h)-sinx]+[cos(x+h)-cosx]}{h}

= \lim_{h\to 0}\frac{[2sin\frac{x+h-x}{2}cos\frac{x+h+x}{2}]+[-2sin\frac{x+h+x}{2}sin\frac{x+h-x}{2}]}{h}

= \lim_{h\to 0}\frac{sinh.cos\frac{2x+h}{2}-2sin(x+\frac{h}{2})sinh}{h}

= \lim_{h\to 0}\frac{sinh}{h}[cos\frac{x+h}{2}-sin(x+\frac{h}{2})]

= cosx - sinx

Question 4. Differentiate each of the following using first principles:

(i) tan²x

Solution:

Given that f(x) = tan²x

By using the formula

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

We get

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

= \lim_{h\to 0}\frac{[tan(x+h)+tanx][tan(x+h)-tanx]}{h}

= \lim_{h\to 0}\frac{\frac{sin(x+h+x)}{cos(x+h)cosx}×\frac{sin(x+h-x)}{cos(x+h)cosx}}{h}

= \lim_{h\to 0}\frac{sin(2x+h)}{cos(x+h)cosx}×\frac{sinh}{cos(x+h)cosx}

= \lim_{h\to 0}\frac{sinh}{h}×\frac{sin2x}{cos^2(x+h)cos^2x}

= \lim_{h\to 0}\frac{sin2x}{cos^2x.cos^2x}

= \lim_{h\to 0}\frac{2sinx.cosx}{cos^2x}×\frac{1}{cos^2x}

= 2tanx sec²x

(ii) tan(2x + 1)

Solution:

Given that f(x) = tan(2x+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{tan[2(x+h)+1]-tan(2x+1)}{h}

= \lim_{h\to 0}\frac{sin(2x+2h+1-2x-1)}{h.cos2(x+h+1)cos(2x+1)}

= \lim_{h\to 0}\frac{2sin2h}{2h.cos(2x+2h+1)cos(2x+1)}

Multiplying both, numerator and denominator by 2.

= \lim_{h\to 0}2(\frac{sin2h}{2})\frac{1}{cos(2x+2h+1)cos(2x+1)}

= \frac{2}{cos^2(2x+1)}

= 2sec²(2x+1)

(iii) tan2x

Solution:

Given that f(x) = tan2x

By using the formula

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

We get

= \lim_{h\to 0}\frac{tan2(x+h)-tan2x}{h}

= \lim_{h\to 0}\frac{sin(2x+2h-2x)}{h.cos(2x+2h)cos2x}

= \lim_{h\to 0}\frac{sin2h}{h.cos(2x+2h)cos2x}

= \lim_{h\to 0}(\frac{sin2h}{2h})×\frac{1×2}{cos(2x+2h)cos2x}

= \frac{2}{cos2x.cos2x}

= 2sec²2x

(iv) √tanx

Solution:

Given that f(x) = √tanx

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{tan(x+h)}-\sqrt{tanx}}{h}

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

We get

= \lim_{h\to 0}\frac{tan(x+h)-tanx}{h(\sqrt{tan(x+h)}+\sqrt{tanx})}

= \lim_{h\to 0}\frac{sin(x+h-x)}{h.cos(x+h)cosx[\sqrt{tan(x+h)}+\sqrt{tanx}]}

= \lim_{h\to 0}\frac{sinh}{h}×\frac{1}{cos(x+h)cosx[\sqrt{tan(x+h)}+\sqrt{tanx}]}

= \lim_{h\to 0}\frac{1}{cos^2x.2\sqrt{tanx}}

= \frac{sec^2x}{2\sqrt{tanx}}

Question 5. Differentiate each of the following using first principles:

(i) sin\sqrt{2x}

Solution:

Given that f(x) = sin\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{sin\sqrt{2(x+h)}-sin\sqrt{2x}}{h}

= \lim_{h\to 0}\frac{2sin[\frac{\sqrt{2(x+h)}-\sqrt{2x}}{2}]cos[\frac{\sqrt{2(x+h)}+\sqrt{2x}}{2}]}{h}

= \lim_{h\to 0}\frac{sin[\frac{\sqrt{2(x+h)}-\sqrt{2x}}{2}]}{[\frac{\sqrt{2(x+h)}-\sqrt{2x}}{2}]}\frac{(\sqrt{2(x+h)}-\sqrt{2x})(\sqrt{2(x+h)}+\sqrt{2x})}{(\sqrt{2(x+h)}+\sqrt{2x})h}cos[\frac{\sqrt{2(x+h)+2x}}{2}]

= \lim_{h\to 0}\frac{sin[\frac{\sqrt{2(x+h)}-\sqrt{2x}}{2}]}{[\frac{\sqrt{2(x+h)}-\sqrt{2x}}{2}]}\lim_{h\to 0}\frac{{2(x+h)}-2x}{(\sqrt{2(x+h)}+\sqrt{2x})h}\lim_{h\to 0}cos[\frac{\sqrt{2(x+h)+2x}}{2}]

= 1×\frac{2}{2\sqrt{2x}}cos(\sqrt{2x})

= \frac{cos\sqrt{2x}}{\sqrt{2x}}

(ii) cos√x

Solution:

Given that f(x) = cos√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{cos\sqrt{x+h}-cos\sqrt{x}}{h}

= \lim_{h\to 0}\frac{-2sin[\frac{\sqrt{x+h}+\sqrt{x}}{2}].sin[\frac{\sqrt{x+h}-\sqrt{x}}{2}]}{h}

= \lim_{h\to 0}\frac{-2sin[\frac{\sqrt{x+h}-\sqrt{x}}{2}][\sqrt{x+h}-\sqrt{x}[\sqrt{x+h}+\sqrt{x}]}{h[\frac{\sqrt{x+h}-\sqrt{x}}{2}][\sqrt{x+h}+\sqrt{x}]}×sin[\frac{\sqrt{x+h}+\sqrt{x}}{2}]

Multiplying numerator and denominator by (\sqrt{x+h}-\sqrt{x})

= \lim_{h\to 0}\frac{-sin[\frac{\sqrt{x+h}-\sqrt{x}}{2}]}{[\frac{\sqrt{x+h}-\sqrt{x}}{2}]}×\frac{x+h-x}{(\sqrt{x+h}+\sqrt{x})h}×sin[\frac{\sqrt{x+h}+\sqrt{x}}{2}]

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

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

= -\frac{sin\sqrt{x}}{2\sqrt{x}}

(iii) tan√x

Solution:

Given that f(x) = tan√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{tan\sqrt{x+h}-tan\sqrt{x}}{h}

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

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

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

= \lim_{h\to 0}\frac{sin(\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}-\sqrt{x)}}×\frac{1}{(\sqrt{x+h}+\sqrt{x})cos\sqrt{x}.cos\sqrt{x+h}}

= 1×\frac{1}{2\sqrt{x}.cos\sqrt{x}.cos\sqrt{x+h}}

= \frac{1}{2\sqrt{x}cos^2x}

= \frac{sec^2\sqrt{x}}{2\sqrt{x}}

(iv) tanx²

Solution:

Given that f(x) = tanx²

By using the formula

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

We get

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

= \lim_{h\to 0}\frac{\frac{sin(x+h)^2}{cos(x+h)^2}-\frac{sinx^2}{cosx^2}}{h}

= \lim_{h\to 0}\frac{\frac{sin(x+h)^2cosx^2-cos(x+h)^2sinx^2}{{cos(x+h)^2}cosx^2}}{h}

= \lim_{h\to 0}\frac{sin((x+h)^2-x^2)}{h.{cos(x+h)^2}.cosx^2}

= \lim_{h\to 0}\frac{sin(x^2+h^2+2hx-x^2)}{h.{cos(x+h)^2}.cosx^2}

= \lim_{h\to 0}\frac{sin(h^2+2hx)}{h.{cos(x+h)^2}.cosx^2}

= \lim_{h\to 0}\frac{sinh}{h}×\frac{(h+2x)}{{cos(x+h)^2}.cosx^2}

= 1.\frac{2x}{cos^2(x)^2}

= 2xsec²x²

Question 6. Differentiate each of the following using first principles:

(i) -x

Solution:

Given that f(x) = -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)+x}{h}

= \lim_{h\to 0}\frac{-x-h+x}{h}

= -1

(ii) (-x)^-1

Solution:

Given that f(x) = (-x)^-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{\frac{-1}{x+h}+\frac{1}{x}}{h}

= \lim_{h\to 0}\frac{-x+x+h}{hx(x+h)}

= \lim_{h\to 0}\frac{1}{x^2+xh}

= 1/x²

(iii) sin(x + 1)

Solution:

Given that f(x) = sin(x+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{sin(x+h+1)-sin(x+1)}{h}

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

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

= \lim_{h\to 0}cos(\frac{2x+2+h}{2})(\frac{sin(h/2)}{h/2})

= cos(\frac{2(x+1)}{2})

= cos(x+1)

(iv) cos(x - π/8)

Solution:

We have, f(x) = cos(x - π/8)

By using the formula

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

We get

= \lim_{h\to 0}\frac{cos(x+h-π/8)-cos(x-π/8)}{h}

= \lim_{h\to 0}-2sin\frac{sin(\frac{2x+h-2π/8}{2})-sin(\frac{h+x-π/8-x+π/8}{2})}{h}

= \lim_{h\to 0}\frac{-2sin(\frac{2x+h-2π/8}{2})sin(h/2)}{2.h/2}

= \lim_{h\to 0}\frac{-2sin(\frac{2x+h-2π/8}{2})}{2}\lim_{h\to 0}\frac{sin(h/2)}{h/2}

= \lim_{h\to 0}sin(\frac{2x+h-2π/8}{2})

= sin(\frac{2x-2π/8}{2})

= -sin(x + π/8)

Read More:

• Derivative of a Function
• Differentiation

Summary

Exercise 30.2 Set 2 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.