KC Sinha: Exercise 26.3- Mathematics Solution Class 11 Chapter 26 अवकलज

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Question 1

x के सापेक्ष निम्नलिखित का अवकल गुणांक निकाले।

[Find the derivatives (differential coefficient) of the following functions w.r.t x]

(i) 10x⁵

Sol :

माना y=10x⁵

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{d x}=\frac{d\left(10 x^{5}\right)}{d x}$

$=\frac{10 d\left(x^{5}\right)}{d x}$

(ii) $2 x-\frac{3}{4}$

Sol :

माना $y=2 x-\frac{3}{4}$

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{dx}=\frac{d(2 a)}{d x}-\frac{d\left(\frac{3}{4}\right)}{dx}$

$\frac{d y}{d x}=2$

(iii) $4 \sqrt{x}-2$

Sol :

माना $y=4 \sqrt{x}-2$

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{d x}=\frac{d(4 \sqrt{x})}{d x}-\frac{d(2)}{dx}$

$=4\times {\frac{1}{2 \sqrt{x}}}-0$

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

(iv) $\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$

Sol :

माना $y=\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$

y=ax^-4-bx^-2+cos x

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{a d\left(x^{-4}\right)}{d x}-\frac{b \cdot d\left(x^{-2}\right)}{d x}+\frac{d(\cos x)}{d x}$

y=a(-4)x^-5-b(-2)x^-3-sin x

$=\frac{-4 a}{x^{5}}+\frac{2 b}{x^{3}}-\operatorname{sin} x$

(iv) $\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$

Sol :

माना y=3cosec x

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{3 \cdot d(\cos e x)}{d x}$

$=-3 \operatorname{cosec x} \cot x$

(vi) $\sqrt{\frac{1+\cos 2 x}{1-\cos 2 x}}$ जहाँ(where)$\pi<x<\frac{3 \pi}{2}$

Sol :

माना $y=\sqrt{\frac{1+\cos 2 x}{1-\cos 2 x}}$

$y=\sqrt{\frac{2 \cos ^{2} x}{2\sin ^2 x}}$

$y=\sqrt{\cot ^{2} x}$

y=cotx

Differentiating w.r.t x

$\frac{d y}{d x}=-\operatorname{cosec}^{2} x$

(vii) $2\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}$

Sol :

माना $y=2\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}$

y=2(1+sin x)

y=2+2sin x

Differentiating w.r.t. x

$\frac{d y}{d x}=d \frac{(2)}{dx}+2 \frac{d(\sin x)}{dx}$

=2cos x

Question 2

यदि(If) $y=\frac{1}{1+x^{n-m}+x^{p-m}}+\frac{1}{1+x^{m-n}+x^{p-n}}+\frac{1}{1+x^{m-p}+x^{n-p}}$ ,.साबित करे कि (Show that) $\frac{d y}{d x}=0$

Sol :

$y=\frac{1}{1+x^{n-m}+x^{p-m}}+\frac{1}{1+x^{m-n}+x^{p-n}}+\frac{1}{1+x^{m-p}+x^{n-p}}$

$y=\frac{1}{1+\frac{x^{n}}{x^{m}}+\frac{x^{p}}{x^{m}}}+\frac{1}{1+\frac{x^{m}}{x^{n}}+\frac{x^{p}}{x^{n}}}+\frac{1}{1+\frac{x^{m}}{x^{p}}+\frac{x^{n}}{x^{p}}}$

$y=\frac{1}{\frac{x^{m}+x^{n}+x^{p}}{x^{m}}}+\frac{1}{\frac{x^{n}+x^{m}+x^{p}}{x^{n}}}+\frac{1}{\frac{x^{p}+x^{m}+x^{n}}{x^p}}$

$y=\frac{x^{m}}{x^{m}+x^{n}+x^{p}}+\frac{a^{n}}{x^{m}+x^{n}+x^{p}}+\frac{x^{p}}{x^{m}+x^{n}+x^{p}}$

$y=\frac{x^{m}+x^{n}+x^{p}}{x^{m}+x^{n}+x^{p}}$

y=1

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(1)}{d x}$

$\frac{d y}{d x}=0$

Question 3

यदि(If) f(x)=xⁿ तथा (and) f '(1)=5 ,साबित करे कि(show that) n=5

Sol :

f(x)=xⁿ

Differentiating w.r.t x

$f^{\prime}(x)=n \cdot x^{n-1}$

x=1,

$f^{\prime}(1)=n(1)^{n-1}$

5=n×1

5=n

Question 4

x के सापेक्ष निम्नलिखित फलनो का अवकल गुणांक निकाले ।

[Find the derivatives(differential coefficient) of the following functions w.r.t x]

(i) 3x⁵+3x³-5

Sol :

माना y=3x⁵+3x³-5

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{3 \cdot d\left(x^{5}\right)}{d x}+3 \cdot \frac{d\left(x^{3}\right)}{d x}-\frac{d(5)}{dx}$

=3×5x⁴+3×x²-0

=15x⁴+9x²

(ii) 3sinx+2sinα, जहाँ α एक अचर है(where α is a constant)

Sol :

माना y=3sinx+2sinα

Differentiating w.r.t. x

$\frac{d y}{d x}=\frac{3 \cdot d\left(\sin x\right)}{dx}+\frac{d(2\sin \alpha)}{d x}$

=3cosx+0

=3cos x 

(iii) (x-2)(x-3)

Sol :

y=(x-2)(x-3)

y=x²-5x+6

Differentiating w.r.t x

$=\frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}-5 \cdot \frac{d(x)}{d x}+\frac{d(6)}{d x}$

=2x-5+0

=2x-5

(iv) cotx+2cosx+3sinx

Sol :

माना y=cotx+2cosx+3sinx

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{d(\cot x)}{dx}+2 \cdot \frac{d(\cos x)}{d x}+3 d \frac{\sin x}{d x}$

=-cosecx²-2sinx+3cosx

(v) $\sqrt{x}-\frac{1}{\sqrt{x}}$

Sol :

माना $y=\sqrt{x}-\frac{1}{\sqrt{x}}$

$y=\sqrt{x}-\frac{1}{x^{1 / 2}}$

$y=\sqrt{x}-x^{-1 / 2}$

Differentiating w.r.t. x

$\frac{d y}{dx}=\frac{d(\sqrt{x})}{dx}-\frac{d\left(x^{-\frac{1}{2}}\right)}{dx}$

$=\frac{1}{2 \sqrt{x}}-\left(\frac{-1}{2}\right) x^{-3 / 2}$

$=\frac{1}{2 \sqrt{x}}+\frac{1}{2 x^{3 / 2}}$

$=\frac{1}{2 \sqrt{x}}+\frac{1}{2 x \cdot \sqrt{x}}$

$=\frac{x+1}{2 x \sqrt{x}}$

$=\frac{x+1}{2 x^{3 / 2}}$

(vi) cosx-2sinx

Sol :

माना y=cosx-2sinx

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(\cos x)}{d x}-2 \cdot \frac{d(\operatorname{sin} x)}{d x}$

=-sinx-2cosx

(vii) x³+7x²+4x+9

Sol :

माना y=x³+7x²+4x+9

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left(x^{3}\right)}{d x}+7 \cdot \frac{d\left(x^{2}\right)}{d x}+4 \cdot \frac{d(x)}{d x}+\frac{d(9)}{d x}$

=3x²+7×2x+4(1)+0

=3x²+14x+4

(viii) 6x¹⁰⁰-x⁵⁵+x

Sol :

माना y=6x¹⁰⁰-x⁵⁵+x

$\frac{d y}{dx}=6 \cdot \frac{d\left(x^{100}\right)}{dx}-\frac{d\left(x^{55}\right)}{d x}+\frac{d(x)}{d x}$

=6×100x⁹⁹-55x⁵⁴+1

=600x⁹⁹-55x⁵⁴+1

(ix) 3cotx+5cosecx

Sol :

माना y=3cotx+5cosecx

$\frac{d y}{dx}=\frac{3 \cdot d(\cos x)}{d x}+5 \cdot d \frac{(\cos x)}{d x}$

=-3cosec²x-5cosecx cotx

(x) 5secx+4cosx

Sol :

$\frac{d y}{dx}=\frac{5d(\tan x)}{d x}+4 \cdot \frac{d(\cos x)}{d x}$

=5secxtanx-4sinx

(xi) 2tanx-7secx

Sol :

माना y=2tanx-7secx

$\frac{d y}{dx}=\frac{2 \cdot d(\tan x)}{dx}-\frac{7d(\operatorname{sec} x)}{d x}$

=2sec²x-7secxtanx

(xii) 5sinx-6cosx+7

Sol :

माना y=5sinx-6cosx+7

Question 5

निम्नलिखित फलनो का x के सापेक्ष अवकलन करे।

[Differentiate the following functions w.r.t. to x]

(i) ax³+bx²+cx+d

Sol :

माना y=ax³+bx²+cx+d

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{d x}=\frac{a \cdot d\left(x^{3}\right)}{d x}+b \cdot \frac{d\left(x^{2}\right)}{d x}+c \frac{d(x)}{dx}+\frac{d(d)}{d x}$

=a3x²+b2x+c×1+0

=3ax²+2bx+c

(ii) x⁴+7x³+8x²+3x+2

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{dx}=\frac{d\left(x^{4}\right)}{d x}+7 \cdot \frac{d\left(x^{3}\right)}{d x}+8 \cdot \frac{d\left(x^{2}\right)}{d x}+3 d \frac{(x)}{d x}+\dfrac{d(2)}{dx}$

=4x³+7×3x²+8×2x+3×1+0

(iii) x³+sinx

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{d x}=\frac{d\left(x^{3}\right)}{d x}+\frac{d(\sin x)}{dx}$

=3x²+cosx

(iv) x³+7x+3+4x²ⁿ+5a²,जहाँ a एक अचर है(where a is constant)

Sol :

माना y=x³+7x+3+4x²ⁿ+5a²

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

=3x²+7+0

(v) xⁿ+ax^n-1+a²x^n-2+..+a^n-1x+aⁿ, जहाँ a एक अचर वास्तविक संख्या है(where a is a fixed real number)

Sol :

y=xⁿ+ax^n-1+a²x^n-2+..+a^n-1x+aⁿ

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{dx}=\frac{d\left(x^{n}\right)}{d x}+a \cdot \frac{d\left(x^{n-1}\right)}{d x}+a^{2} \cdot \frac{d\left(x^{n-z}\right)}{d x}+\ldots+a^{n-1} \cdot \frac{d(x)}{dx}+\frac{d(a^n)}{dx}$

=nx^n-1+a(n-1)x^n-2+a²(n-2)x^n-3+..+a^n-1

(vi) a_nxⁿ+a_n-1x^n-1+...+a₁x+a₀, जहाँ सभी a₁ एक अचर है तथा a_n≠0 (where a₁ are constant and a_n≠0)

Sol :

माना y=a_nxⁿ+a_n-1x^n-1+...+a₁x+a₀

Differentiating w.r.t x

x के सापेक्ष अवकलन करने पर

$\frac{d y}{dx}=a_{n} \cdot \frac{d\left(x^{n}\right)}{dx}+a_{n-1} \cdot \frac{d\left(x^{4-1}\right)}{dx}+\cdots \cdot \cdot a_1 \frac{d(x)}{dx}+\frac{d(a_0)}{dx}$

=a_nnx^n-1+a_n-1(n-1)x^n-2+...+a₁

Question 6

निम्नलिखित फलनो का x के सापेक्ष अवकलन करे।

[Find the d.c. of the following functions w.r.t x]

(i) $\frac{5 x^{4}+6 x^{2}-x+1}{x}$

Sol :

माना $y=\frac{5 x^{4}+6 x^{2}-x+1}{x}$

$y=\frac{5 x^{4}}{x}+\frac{6 x^{2}}{x}-\frac{x^{1}}{x}+\frac{1}{x}$

$y=5 x^{3}+6 x-1+\frac{1}{x}$

Differentiating w.r.t x

$\frac{d y}{d x}=5 \cdot \frac{d\left(x^{3}\right)}{d x}+6 \cdot \frac{d(x)}{d x}-\frac{d(1)}{dx}+\frac{d\left(\frac{1}{x}\right)}{dx}$

$=5 \times 3 x^{2}+6 \times 1-0-\frac{1}{x^{2}}$

$=15 x^{2}+6-\frac{1}{x^{2}}$

(ii) $\frac{(x+5)\left(2 x^{2}-1\right)}{x}$

Sol :

माना $y=\frac{(x+5)\left(2 x^{2}-1\right)}{x}$

$y=\frac{2 x^{3}-x+10 x^{2}-5}{x}$

$y=\frac{2 x^{3}}{x}-\frac{x}{x}+\frac{10 x^{2}}{x}-\frac{5}{x}$

$y=2 x^{2}-1+10 x-\frac{5}{2}$

Differentiating w.r.t x

$\frac{d y}{dx}=2 \cdot \frac{d\left(x^{2}\right)}{dx}-\frac{d(1)}{dx}+10 \cdot \frac{d(x)}{dx}-5 \cdot \frac{d\left(\frac{1}{x}\right)}{d x}$

$=2 \times 2 x-0+10-5\left(\frac{-1}{x^{2}}\right)$

$=4 x+10+\frac{5}{x^{2}}$

(iii) $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}$

sol :

$y=\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}$

$y=(\sqrt{x})^{2}+\left(\frac{1}{\sqrt{x}}\right)^{2}+2 \cdot \sqrt{x} \cdot \frac{1}{\sqrt{x}}$

$y=x+\frac{1}{x}+2$

Differentiating w.r.t x

$\frac{d y}{d x}=1-\frac{1}{x^{2}}$

(iv) $\frac{x^{2} \tan x+1}{\tan x}$

Sol :

$y=\frac{x^{2} \tan x+1}{\tan x}$

$=\frac{x^{2} \tan x}{\tan x}+\frac{1}{\tan x}$

y=x²+cotx

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d\left(x^{2}\right)}{dx}+d \frac{(\cot x)}{d x}$

=2x-cosec²x

(v) (x-a)(x-b), जहाँ a और b अचर है(where a and b are constant)

Sol :

माना y=(x-a)(x-b)

y=x²-bx-ax+ab

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d\left(x^{2}\right)}{d x}-b \cdot \frac{d(x)}{dx}-a \cdot \frac{d(x)}{dx}+\frac{d(a s)}{dx}$

=2x-b-a+0

=2x-a-b

(vi) (ax²+b)²,जहाँ a और b अचर है(where a and b are constant)

Sol :

माना y=(ax²+b)²

y=(ax²)²+2.ax²+b+b²

y=a²x⁴+2abx²+b²

Differentiating w.r.t x

$\frac{d y}{d x}=a^{2} \cdot d \frac{\left(x^{4}\right)}{dx}+\frac{2 a b d\left(x^{2}\right)}{dx}+\frac{d\left(b^{2}\right)}{dx}$

=a²4x³+2ab×2x+0

=4ax(ax²+b)

Question 7

$\frac{d y}{d x}$ निकाले(Find $\frac{d y}{d x}$ when)

(i) $y=8 x^{2}-x+5-\frac{3}{x^{3}}$

Sol :

Differentiating w.r.t x

$\frac{d y}{dx}=8 \cdot \frac{d\left(x^{2}\right)}{d x}-\frac{d(x)}{dx}+\frac{d(5)}{d}-3 d \frac{\left(x^{-3}\right)}{dx}$

=8×2x-1+0-3(-3)x^-4

=16x-1$+\frac{9}{x^4}$

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

Sol :

$y=\frac{2 x^{2}-3 x+1}{x^{1 / 2}}$

$y=\frac{2 x^{2}}{x^{1 / 2}}-\frac{3 x}{x^{\frac{1}{2}}}+\frac{1}{x^{1 / 2}}$

$y=2 x^{3 / 2}-3 x^{1 / 2}+x^{-\frac{1}{2}}$

Differentiating w.r.t x

$\frac{dy}{dx}=2 \frac{d\left(x^{3 / 2}\right)}{dx}-3 \frac{d\left(x^{\frac{1}{2}}\right)}{dx}+d \frac{\left(x^{-\frac{1}{2}}\right)}{dx}$

$=2 \cdot\left(\frac{3}{2}\right) x^{\frac{1}{2}}-3\left(\frac{1}{2}\right) x^{-1 / 2}+\left(\frac{-1}{2}\right)x^{-3 / 2}$

$=\frac{6}{2} x^{\frac{1}{2}} \cdot \frac{-3}{2 x^{1 / 2}}-\frac{1}{2 x^{3 / 2}}$

$=\frac{6 x^{2}-3 x-1}{2 x^{3 / 2}}$

(iii) $y=\frac{x^{3}+5 x^{2}+x+7}{x}$

Sol :

$y=\left(x^{2}\right)^{3}+3 \cdot\left(x^{2}\right)^{2} \cdot \frac{1}{x^{2}}+3 \cdot x^{2} \cdot\left(\frac{1}{x^{2}}\right)^{2}+\left(\frac{1}{x^{2}}\right)^{3}$

$y=x^{6}+3 x^{2}+\frac{3}{x^{2}}+\frac{1}{x^{6}}$

$y=x^{6}+3 x^{2}+3 x^{-2}+x^{-6}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left(x^{6}\right)}{dx}+\frac{3 \cdot d\left(x^{2}\right)}{dx}+3 \cdot \frac{d\left(x^{-2}\right)}{dx}+\frac{d\left(x^{-6}\right)}{d x}$

=6x⁵+3×2x+3(-2)x^-3+(-6)x^-7

$=6 x^{5}+6 x-\frac{6}{x^{3}}-\frac{6}{x^{7}}$

(iv) $y=\left(x^{2}+\frac{1}{x^{2}}\right)^{3}$

Sol :

(v) $y=\left(x-\frac{1}{x}\right)\left(x^{2}+\frac{1}{2 x}\right)$

Sol :

$y=\left(x-\frac{1}{2}\right)\left(x^{2}+\frac{1}{2 x}\right)$

$y=x^{3}+\frac{1}{2}-x-\frac{1}{2 x^{2}}$

$y=x^{3}+\frac{1}{2}-x-\frac{1}{2} x^{-2}$

Differentiating w.r.t x

$\frac{d y}{d x}=3 x^{2}+0-1-\frac{1}{2} \cdot(-2) \cdot x^{-3}$

$=3 x^{2}-1+\frac{1}{x^{3}}$

(vi) $y=\left(x+\frac{1}{x}\right)^{3}$

Sol :

$y=x^{3}+3 \cdot x^{2} \cdot \frac{1}{x}+3 \cdot x \cdot\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{x}\right)^{3}$

Differentiating w.r.t x

$\frac{d y}{d x}=3 x^{2}+3+3\left(-\frac{1}{x^{2}}\right)+(-3) x^{-4}$

$=3 x^{2}+3-\frac{3}{x^{2}}-\frac{3}{x^{4}}$

(vii) $y=\frac{a+b \cos x}{\sin x}$

Sol :

$y=\frac{a}{\sin x}+b \frac{\cos x}{\sin x}$

y=acosecx+bcotx

Differentiating w.r.t x

$\frac{d y}{d n}=-a \operatorname{cosec} x \operatorname{cot} x-b \cos x c^{2} x$

(viii) $y=\frac{b+c \tan x}{\sec x}$

Sol :

$y=b \cos x+c \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}$

y=bcosx+csinx

Differentiating w.r.t x

$\frac{d y}{dx}=-b \sin x+c \cos x$

(ix) x⁵(3-6x^-9)

Sol :

y=3x⁵-6x^-4

Differentiating w.r.t x

$\frac{d y}{dx}=3 \times 5 x^{4}-6(-4) \cdot x^{-5}$

$=15 x^{4}+\frac{24}{x^{5}}$

(x) x^-3(5+3x)

Sol :

(xi) x^-4(3-4x^-5)

Sol :

y=x^-4(3-4x^-5)

y=3x^-4-4x^-9

Differentiating w.r.t x

$\frac{d y}{dx}=3 \cdot(-4) x^{-5}-4(-9) x^{-10}$

$=\frac{-12}{x^{5}}+\frac{36}{x^{10}}$

(xii) (5x³+3x-1)(x-1)

Sol :

Question 8

वक्र f(x)=2x⁶+x⁴-1 का x=1 पर स्पर्श रेखा की ढाल निकाले ।

[Find the slope of the tangent to the curve f(x)=2x⁶+x⁴-1 at x=1]

Sol :

f(x)=2x⁶+x⁴-1 

Differentiating w.r.t x

स्पर्श रेखा की ढाल (x=1)=f '(1)

=12(1)⁵+4(1)³

=12+4

=16

Question 9

यदि f(x)=x²-9x+20 तो f '(x) निकाले और इससे f '(100) तथा $f^{\prime}\left(\frac{9}{2}\right)$ निकाले

Sol :

f(x)=x²-9x+20

Differentiating w.r.t x

f '(x)=2x-9

f '(100)=2(100)-9

=200-9

=191

$f^{\prime}\left(\frac{9}{2}\right)=2\left(\frac{9}{2}\right)-9=0$

Question 10

यदि (If) $y=\frac{2-3 \cos x}{\sin x}, x=\frac{\pi}{4}$ पर $\frac{d y}{d x}$ निकाले (find $\frac{d y}{d x}$ at $\left.x=\frac{\pi}{4}\right)$

Sol :

$y=\frac{2-3 \cos x}{\sin x}$

$y=\frac{2}{\sin x}-\frac{3 \cos x}{\operatorname{sin} x}$

y=2cosec x-3cot x

Differentiating w.r.t x

$\frac{dy}{dx}=-2 \operatorname{cosec} x \cot x+3 \operatorname{cosec}^{2} x$

$x=\frac{\pi}{4}$ पर,

$\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=-2 \operatorname{cosec} \frac{\pi}{4} \cot \frac{\pi}{4}+3 \operatorname{cosec}^{2} \frac{\pi}{4}$

$=-2 \times \sqrt{2} \times 1+3(\sqrt{2})^{2}$

$=-2 \sqrt{2}+6=6-2 \sqrt{2}$

Question 11

यदि y=(x-a)(x-b) तो x का मान निकाले जिसके लिए $\frac{d y}{d x}=0$

Sol :

y=x²-bx-ax+ab

Differentiating w.r.t x

$\frac{dy}{dx}=2 x-b-a+0$

$\frac{dy}{dx}=2 x-(a+b)$

$\because \frac{d y}{d x}=0$

2x-(a+b)=0

2x=a+b

$x=\frac{a+b}{2}$

Question 12

यदि (If) $f(x)=\frac{x-4}{2 \sqrt{x}}$ , तो f '(4) निकााले (then f '(4))

Sol :

$f(x)=\frac{x-4}{2 x^{1 / 2}}$

$=\frac{x}{2 x^{1 / 2}}-\frac{4}{2 x^{1 / 2}}$

$f(x)=\frac{1}{2} \sqrt{x}-2 x^{-1/ 2}$

Differentiating w.r.t x

$f^{\prime}(x)=\frac{1}{2} \times \frac{1}{2 \sqrt{x}}-2 \cdot\left(\frac{-1}{2}\right) x^{-3 / 2}$

$f^{\prime}(x)=\frac{1}{4 \sqrt{x}}+\frac{1}{x^{\frac{3}{2}}}$

$f^{\prime}(x)=\frac{1}{4 \sqrt{x}}+\frac{1}{x \sqrt{2}}$

x=4 , 

$f^{\prime}(4)=\frac{1}{4 \sqrt{4}}+\frac{1}{4 \sqrt{4}}$

$=\frac{1}{8}+\frac{1}{8}$

$=\frac{2}{8}=\frac{1}{4}$

Question 13

यदि (If) f(x)=1+x+x²+x³+...x⁵⁰ , तो f '(1) निकाले (then find f '(1) )

Sol :

f(x)=1+x+x²+x³+...x⁵⁰

Differentiating w.r.t x

f '(x)=1+2x+3x²+...50x⁴⁹

x=1 पर,

f '(x)=1+2(1)+3(1)²+...50(1)⁴⁹

=1+2+3+...50

$=\frac{50 (50+1)}{2}$

=25×51

=1275

Question 14

यदि(If) $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1$, साबित केर कि(prove that) f '(1)=100f '(0)

Sol :

$f(x)=\frac{x^{100}}{(100)}+\frac{x^{99}}{99}+\dots+\frac{x^{2}}{2}+x+1$

Differentiating w.r.t x

$f^{\prime}(x)=\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\dots \frac{2 x}{2}+1+0$

f '(x)=x⁹⁹+x⁹⁸+...+x+1

f '(x)=(1)⁹⁹+(1)⁹⁸+....+1+1=100

100f '(0)=100[(0)⁹⁹+(0)⁹⁸+....0+1]

=100×1

=100

∴f '(1)=100f '(0)

Question 15

यदि (If) y=cosecx+cotx, साबित करे कि(prove that) $\frac{d y}{d x}+y \operatorname{cosec} x=0$

Sol :

y=cosecx+cotx

Differentiating w.r.t x

$\frac{d y}{d x}=-\operatorname{cosec} x \cot x-\operatorname{cosec}^{2} x$

$\frac{d y}{dx}=-\operatorname{cosec} x(\cot x+\operatorname{cosec} x)$

$\frac{dy}{dx}=-y \operatorname{cosec} x$

$\frac{d y}{d x}+y \operatorname{cosec} x=0$

Question 16

यदि (If) $y=x+\frac{1}{x}$, साबित करे कि (prove that) $x^{2} \frac{d y}{d x}-x y+2=0$

Sol :

$y=x+\frac{1}{x}$

Differentiating w.r.t x

$\frac{d y}{d x}=1-\frac{1}{x^{2}}$

L.H.S

$x^{2} \frac{dy}{d x}-x y+2$

$=x^{2}\left(1-\frac{1}{x^2}\right)-x\left(x+\frac{1}{x}\right)+2$

=x²-1-x²-1+2

=-2+2

=0

Question 17

यदि(If) $y=\sqrt{x}+\frac{1}{\sqrt{x}}$ , साबित करे कि (show that) $2 x \frac{d y}{d x}+y=2 \sqrt{x}$

Sol :

$y=\sqrt{x}+\frac{1}{\sqrt{x}}$

$y=\sqrt{x}+\frac{1}{x^{1 / 2}}$

$y=\sqrt{x}+x^{-\frac{1}{2}}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}+\left(-\frac{1}{2}\right) x^{-3 / 2}$

$\frac{d y}{dx}=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{3} / 2}$

$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x \sqrt{x}}$

L.H.S

$2 x \frac{d y}{dx}+y$

$=2 x\left(\frac{1}{2 \sqrt{x}}-\frac{1}{2 x \sqrt{x}}\right)+\sqrt{x}+\frac{1}{\sqrt{x}}$

$=\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}$

$=2 \sqrt{x}$=R.H.S