RD Sharma Class 12 Ex 11.1 Solutions Chapter 11 Differentiation

Here we provide RD Sharma Class 12 Ex 11.1 Solutions Chapter 11 Differentiation for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 11.1 Solutions Chapter 11 Differentiation book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter11
Exercise11.1
CategoryRD Sharma Solutions

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 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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