Class 12 RD Sharma Solutions - Chapter 11 Differentiation - Exercise 11.1

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}]}

Summary

Exercise 11.1 focuses on applying differentiation techniques to various types of functions. It covers the differentiation of algebraic, trigonometric, exponential, and logarithmic functions. Students are expected to use basic differentiation rules, chain rule, product rule, and quotient rule. The exercise also includes problems involving implicit differentiation and higher-order derivatives.