Exercise 26.3
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}
=\frac{10 d\left(x^{5}\right)}{d x}
=10×5x4
=50x4
=50x4
(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
Differentiating w.r.t 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
Differentiating w.r.t x
\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 :
माना y=5secx+4cosx
Differentiating w.r.t x
\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
Differentiating w.r.t x
\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
Differentiating w.r.t x
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
Sol :
माना y=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
=4x3+21x2+16x+3
(iii) x3+sinx
Sol :
माना y=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
=3x2+7
(v) xn+axn-1+a2xn-2+..+an-1x+an,
जहाँ a एक अचर वास्तविक संख्या है(where a is a fixed real number)
Sol :
y=xn+axn-1+a2xn-2+..+an-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}
=nxn-1+a(n-1)xn-2+a2(n-2)xn-3+..+an-1
(vi) anxn+an-1xn-1+...+a1x+a0, जहाँ सभी a1 एक अचर है
तथा an≠0 (where a1 are constant and an≠0)
Sol :
माना y=anxn+an-1xn-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}
=annxn-1+an-1(n-1)xn-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
f '(x)=2×6x5+4x3-0
f '(x)=12x5+4x3
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
It's Nice.
ReplyDeleteThanks.