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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.1
  • Last Updated : 07 Apr, 2021

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 e3x

Solution:

We have,



Let,

f(x)=e3x

f(x+h)=e3(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})]

=3e3x

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

Solution:

We have,

Let,

f(x)=eax+b

f(x+h)=ea(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})]

=aeax+b

Question 4. Differentiate the following functions from first principles ecosx

Solution:

We have,

Let,

f(x)=ecosx

f(x+h)=ecos(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}

=ecosx(-sinx)

=-sinx.ecosx

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 x2ex

Solution:

We have,

Let,

f(x)=x2ex

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

=x2ex+2xex+0

=ex(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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