RD Sharma Class 12 Ex 11.1 Solutions Chapter 11 Differentiation

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	11
Exercise	11.1
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 11.1 Solutions Chapter 11 Differentiation

Question 1. Differentiate the following functions from first principles e^-x

Solution:

We have,

Let,

f(x)=e^-x

f(x+h)=e^-(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{-(x+h)}-e^{-x}}{h}$

$=\lim_{h\to0}\frac{e^{-x}.e^{-h}-e^{-x}}{h}$

$=\lim_{h\to0}[e^{-x}(\frac{e^{-h}-1}{-h})](-1)$

$=-e^{-x}[\lim_{h\to0}(\frac{e^{-h}-1}{-h})]$

=-e^-x

Question 2. Differentiate the following functions from first principles e^3x

Solution:

We have,

Let,

f(x)=e^3x

f(x+h)=e^3(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{3(x+h)}-e^{3x}}{h}$

$=\lim_{h\to0}\frac{e^{3x}.e^{3h}-e^{3x}}{h}$

$=\lim_{h\to0}[e^{3x}(\frac{e^{3h}-1}{3h})](3)$

$=3e^{3x}[\lim_{h\to0}(\frac{e^{3h}-1}{3h})]$

=3e^3x

Question 3. Differentiate the following functions from first principles e^ax+b

Solution:

We have,

Let,

f(x)=e^ax+b

f(x+h)=e^a(x+h)+b

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{a(x+h)+b}-e^{ax+b}}{h}$

$=\lim_{h\to0}\frac{e^{ax+b}.e^{ah}-e^{ax+b}}{h}$

$=\lim_{h\to0}[e^{ax+b}(\frac{e^{ah}-1}{ah})](a)$

$=ae^{ax+b}[\lim_{h\to0}(\frac{e^{ah}-1}{ah})]$

=ae^ax+b

Question 4. Differentiate the following functions from first principles e^cosx

Solution:

We have,

Let,

f(x)=e^cosx

f(x+h)=e^cos(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{cos(x+h)}-e^{cosx}}{h}$

$=\lim_{h\to0}[e^{cosx}(\frac{e^{cos(x+h)-cosx}-1}{h})]$

$=e^{cosx}\lim_{h\to0}(\frac{e^{cos(x+h)-cosx}-1}{cos(x+h)-cosx})(\frac{cos(x+h)-cosx}{h})$

$=e^{cosx}\lim_{h\to0}(\frac{cos(x+h)-cosx}{h})$

$=e^{cosx}\lim_{h\to0}(\frac{-2sin(\frac{x+h+x}{2})sin(\frac{x+h-x}{2})}{h})$

$=e^{cosx}\lim_{h\to0}\frac{-2sin(\frac{2x+h}{2})}{2}\frac{sin\frac{h}{2}}{\frac{h}{2}}$

$=e^{cosx}\lim_{h\to0}\frac{-2sin(\frac{2x+h}{2})}{2}\frac{1}{2}$

=e^cosx(-sinx)

=-sinx.e^cosx

Question 5. Differentiate the following functions from first principles e^√2x

Solution:

We have,

Let,

f(x)=e^√2x

f(x+h)=e^√2(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h}$

$=\lim_{h\to0}[e^{\sqrt{2x}}(\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{h})]$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{e^{\sqrt{2(x+h)}-\sqrt{2x}}-1}{\sqrt{2(x+h)}-\sqrt{2x}})(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2(x+h)-2x}{h(\sqrt{2(x+h)}-\sqrt{2x})})$ (After rationalising the numerator)

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2x+2h-2x}{h(\sqrt{2(x+h)}-\sqrt{2x})})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2h}{h(\sqrt{2(x+h)}-\sqrt{2x})})$

$=e^{\sqrt{2x}}\lim_{h\to0}(\frac{2}{(\sqrt{2(x+h)}-\sqrt{2x})})$

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

Question 6. Differentiate the following functions from first principles log(cosx)

Solution:

We have,

Let,

f(x)=log(cosx)

f(x+h)=log(cos(x+h))

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{log(cos(x+h))-log(cosx)}{h}$

$=\lim_{h\to0}\frac{log\frac{cos(x+h)}{cosx}}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{cos(x+h)}{cosx}-1]}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{cos(x+h)}{cosx}]}{\frac{cos(x+h)-cosx}{cosx}}×\lim_{h\to0}\frac{cos(x+h)-cosx}{cosx}$

$=1×\lim_{h\to0}\frac{cos(x+h)-cosx}{h×cosx}$

$=\lim_{h\to0}\frac{-2sin(\frac{x+h+x}{2})sin(\frac{x+h-x}{2})}{h×cosx}$

$=-2\lim_{h\to0}\frac{sin(\frac{2x+h}{2})sin(\frac{h}{2})}{2(\frac{h}{2})×cosx}$

Since, $\lim_{h\to0}\frac{sinx}{x}=1$

=-(2sinx)/(2cosx)

=-tanx

Question 7. Differentiate the following functions from first principles e^√cotx

Solution:

We have,

Let,

f(x)=e^√cotx

f(x+h)=e^√cot(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{e^{\sqrt{cot(x+h)}}-e^{\sqrt{cotx}}}{h}$

$=\lim_{h\to0}[e^{\sqrt{cotx}}(\frac{e^{\sqrt{cot(x+h)}-\sqrt{cotx}}-1}{h})]$

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{e^{\sqrt{cot(x+h)}-\sqrt{cotx}}-1}{\sqrt{cot(x+h)}-\sqrt{cotx}})(\frac{\sqrt{cot(x+h)}-\sqrt{cotx}}{h})$

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{\sqrt{cot(x+h)}-\sqrt{cotx}}{h})$

since, $\lim_{h\to0}\frac{e^x-1}{x}=1$

$=e^{\sqrt{cotx}}\lim_{h\to0}{\frac{cot(x+h)-cotx}{h(\sqrt{cot(x+h)}-\sqrt{cotx})}}$ (After rationalising the numerator)

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{\frac{cot(x+h)cotx+1}{cot(x-x-h)}}{h(\sqrt{cot(x+h)}-\sqrt{cotx})})$

$=e^{\sqrt{cotx}}\lim_{h\to0}(\frac{cot(x+h)cotx+1}{hcot(-h)(\sqrt{cot(x+h)}-\sqrt{cotx})})$

$=-e^{\sqrt{cotx}}\lim_{h\to0}(\frac{cot(x+h)cotx+1}{\frac{h}{tanh}(\sqrt{cot(x+h)}-\sqrt{cotx})})$

Since, $\lim_{h\to0}\frac{tanx}{x}=1$

$=\frac{e^{\sqrt{cotx}}×(cot^2x+1)}{2\sqrt{cotx}}$

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

Question 8. Differentiate the following functions from first principles x²e^x

Solution:

We have,

Let,

f(x)=x²e^x

f(x+h)=(x+h)²e^(x+h)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{(x+h)^2e^{(x+h)}-x^2e^x}{h}$

$=\lim_{h\to0}(\frac{x^2e^{x+h}-x^2e^x}{h}+\frac{2hxe^{x+h}}{h}+\frac{h^2e^{x+h}}{h})$

$=\lim_{h\to0}(\frac{x^2e^x(e^h-1)}{h}+2xe^{x+h}+{he^{x+h}})$

Since, $\frac{e^h-1}{h}=1$

=x²e^x+2xe^x+0

=e^x(x²+2x)

Question 9. Differentiate the following functions from first principles log(cosecx)

Solution:

We have,

Let,

f(x)=log(cosecx)

f(x+h)=log(cosec(x+h))

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{log(cosec(x+h))-log(cosecx)}{h}$

$=\lim_{h\to0}\frac{log\frac{cosec(x+h)}{cosecx}}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{cosec(x+h)}{cosecx}-1]}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{sinx}{sin(x+h)}-1]}{h}$

$=\lim_{h\to0}\frac{log[1+\frac{sinx-sin(x+h)}{sin(x+h)}]}{\frac{sinx-sin(x+h)}{sin(x+h)}}×\lim_{h\to0}\frac{\frac{sinx-sin(x+h)}{sin(x+h)}}{h}$

$=\lim_{h\to0}\frac{sinx-sin(x+h)}{sin(x+h)}$

$=\lim_{h\to0}\frac{2cos(\frac{x+x+h}{2})sin(\frac{x-x-h}{2})}{hsin(x+h)}$

$=\lim_{h\to0}\frac{2cos(\frac{2x+h}{2})sin(\frac{x-x-h}{2})}{sin(x+h)}\frac{sin(-\frac{h}{2})}{-\frac{h}{2}.(-2)}$

=-cotx

Question 10. Differentiate the following functions from first principles sin^-1(2x+3)

Solution:

We have,

Let,

f(x)=sin^-1(2x+3)

f(x+h)=sin^-1[2(x+h)+3]

f(x+h)=sin^-1(2x+2h+3)

$\frac{d}{dx}f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\to0}\frac{sin^{-1}(2x+2h+3)-sin^{-1}(2x+3)}{h}$

$=\lim_{h\to0}\frac{sin^{-1}[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]}{h}$

$=\lim_{h\to0}\frac{sin^{-1}t}{t}\frac{t}{h}$

Where $t=[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]$

$=\lim_{h\to0}\frac{t}{h}$

$=\lim_{h\to0}\frac{[(2x+2h+3)\sqrt{1-(2x+3)^2}-(2x+3)\sqrt{1-(2x+2h+3)^2}]}{h}$

$=\lim_{h\to0}\frac{[(2x+2h+3)^2[1-(2x+3)^2]-(2x+3)^2[1-(2x+2h+3)^2]}{h[(2x+2h+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+2h+3)^2]}}$ (After rationalising the numerator)

Solving above equation

$=\lim_{h\to0}\frac{4h[h+(2x+3)]}{h[(2x+2h+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+2h+3)^2]}}$

$=\frac{4(2x+3)}{[(2x+3)\sqrt{1-(2x+3)^2}+(2x+3)\sqrt{1-(2x+3)^2]}}$

$=\frac{4(2x+3)}{2[(2x+3)\sqrt{1-(2x+3)^2}]}$

$=\frac{2}{2[\sqrt{1-(2x+3)^2}]}$

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RD Sharma Class 12 Ex 11.1 Solutions Chapter 11 Differentiation

Question 1. Differentiate the following functions from first principles e-x

Question 2. Differentiate the following functions from first principles e3x

Question 3. Differentiate the following functions from first principles eax+b

Question 4. Differentiate the following functions from first principles ecosx

Question 5. Differentiate the following functions from first principles e√2x

Question 6. Differentiate the following functions from first principles log(cosx)

Question 7. Differentiate the following functions from first principles e√cotx

Question 8. Differentiate the following functions from first principles x2ex

Question 9. Differentiate the following functions from first principles log(cosecx)

Question 10. Differentiate the following functions from first principles sin-1(2x+3)

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Question 1. Differentiate the following functions from first principles e^-x

Question 2. Differentiate the following functions from first principles e^3x

Question 3. Differentiate the following functions from first principles e^ax+b

Question 4. Differentiate the following functions from first principles e^cosx

Question 5. Differentiate the following functions from first principles e^√2x

Question 7. Differentiate the following functions from first principles e^√cotx

Question 8. Differentiate the following functions from first principles x²e^x

Question 10. Differentiate the following functions from first principles sin^-1(2x+3)