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## S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(a)

Differentiate with respect to x :

Question 1.

(i) x^{5}

(ii) 6x^{8}

(iii) x^{3/4}

(iv) 4\(\sqrt{x}\)

(v) 8x^{-3/4}

(vi) \(\sqrt{x^3}\)

(vii) \(\frac{9}{x}\)

(viii) \(\frac{7}{x^2}\)

Solution:

(i) Let y = x^{5},

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\frac{d}{d x} x^5=5 x^4\)

(ii) Let y = 6x^{8},

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\frac{d}{d x}\left(6 x^8\right)=48 x^7\)

(iii) Let y = x^{3/4}

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\frac{d}{d x} x^{3 / 4}=\frac{3}{4} x^{\frac{3}{4}-1}=\frac{3}{4} x^{\frac{-1}{4}}\)

(iv) Let y = 4\(\sqrt{x}\)

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\frac{d}{d x} 4 \sqrt{x}=4 \frac{d}{d x} x^{1 / 2}\)

= \(4 \times \frac{1}{2} x^{\frac{1}{2}-1}=\frac{2}{\sqrt{x}}\)

(v) Let y = 8x^{-3/4} ;

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=8\left(\frac{-3}{4}\right) x^{\frac{-3}{4}-1}=-6 x^{-7 / 4}\)

(vi) Let y = \(\sqrt{x^3}=x^{3 / 2}\)

Diff both sides w.r.t x, we have

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

(vii) Let y = \(\frac{9}{x}\)

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{9}{x}\right)=9 \frac{d}{d x} \frac{1}{x}\)

= 9\(\frac{d}{d x} x^{-1}=9(-1) x^{-1-1}\)

= \(\frac{-9}{x^2}\)

(viii) Let y = \(\frac{7}{x^2}\) ;

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=7 \frac{d}{d x} x^{-2}=7(-2) x^{-2-1}=\frac{-14}{x^3}\)

Question 2.

(i) (2x + 3)^{5}

(ii) (1 – x)^{4}

(iii) \(\sqrt{8-7 x}\)

(iv) (3x² + 5)^{9}

Solution:

Question 3.

(i) \(\frac{2}{x}+\frac{1}{\sqrt{x}}\)

(ii) (2x – 1) (3x + 2)

(iii) \(x^4-2 x+\frac{1}{x^2}\)

(iv) 2x² (x + 1) + 2

(v) \(\frac{3 x^4-x}{x^3}\)

Solution:

Question 4.

(i) cos 7x

(ii) tan ax

(iii) sec 9x

(iv) sin x²

(v) cos \(\sqrt{x}\)

(vi) 2 cosec bx²

Solution:

(i) Let y = cos 7x ;

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\frac{d}{d x} \cos 7 x\)

= sin 7x \(\frac { d }{ dx }\) (7x) = – 7sin 7x

(ii) Let y = tan ax ;

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\sec ^2 a x \frac{d}{d x} a x\) = a sec²ax

(iii) Let y = sec 9x ;

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\frac{d}{d x}\)sec 9x = sec9x tan9x \(\frac { d }{ dx }\) 9x

= 9 sec 9x tan 9x

(iv) Let y = sin x² ;

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=\cos x^2 \frac{d}{d x} x^2=2 x \cos x^2\)

(v) Let y = cos \(\sqrt{x}\) ;

Diff both sides w.r.t x, we have

\(\frac{d y}{d x}=-\sin \sqrt{x} \frac{d}{d x} x^{\frac{1}{2}}\)

= – sin \(\sqrt{x} \frac{1}{2} x^{\frac{-1}{2}}\)

= – \(\frac{\sin \sqrt{x}}{2 \sqrt{x}}\)

(vi) Let y = 2 cosec bx³ ;

Diff both sides w.r.t A, we have

\(\frac { dy }{ dx }\) = – 2 cosec bx³ cot bx³ \(\frac { d }{ dx }\) bx³

= – 6bx² cosec bx³ cot bx³

Question 5.

(i) \(\frac{x^3}{3 x-2}\)

(ii) \(\frac{x}{\sin x}\)

(iii) \(\frac{1+\cos x}{1-\cos x}\)

(iv) \(x^2-\sqrt{(1+x)}\)

(v) sin 2x cos² x

(vi) tan^{4} 7x

(vii) \(\frac{1}{\sin x+\cos x}\)

(viii) \(\frac{1+\cos x}{x}\)

Solution:

Question 6.

Given that y = \(\frac{\sin x-\cos x}{\sin x+\cos x}\), show that \(\frac{d y}{d x}=1+y^2\)

Solution:

Question 7.

Differentiate with respect to x :

(i) (2x² – 1) (x³ + 4)³

(ii) \(\frac{x^4+1}{\sqrt{1+x}}\)

(iii) \(\left(x+\frac{1}{x}\right)^{-1}\)

(iv) tan^{4}2x

Solution:

Question 8.

Find the gradient function \(\frac { dy }{ dx }\) for each of the following :

(i) y = x – 7x²

(ii) y = 4x^{7} – 3x³ + 5x – 11.

Solution:

(i) Given y = x – 7x² ;

Diff both sides w.r.t x, we have

\(\frac { dy }{ dx }\) = 1 – 14x

(ii) Given y = 4x^{7} – 3x³ + 5x – 11

Diff both sides w.r.t x, we have

\(\frac { dy }{ dx }\) = 28x^{6} – 9x² + 5

Question 9.

Find the gradients of the following curves at the points indicated.

(i) y = x² + 5x at (0, 0)

(ii) y = (x + 1) (2x + 3) at (2, 21)

(iii) y = 2x² – x + \(\frac { 4 }{ x }\) at (2, 8)

Solution:

(i) Given y = x² + 5x ;

Diff both sides w.r.t x, we have

\(\frac { dy }{ dx }\) = 2x + 5

at (0, 0); \(\frac { dy }{ dx }\) = 2 x o + 5 = 5

(ii) Given y = (x + 1) (2x + 3) = 2x² + 5x + 3

Diff both sides w.r.t x, we have

\(\frac { dy }{ dx }\) = 4x + 5

∴ at (2, 21); \(\frac { dy }{ dx }\) = 4 x 2 + 5 = 13

(iii) Given y = 2x² – x + \(\frac { 4 }{ x }\)

Diff both sides w.r.t x, we have

at (2, 8); \(\frac { dy }{ dx }\) = 8 – 1 – \(\frac { 4 }{ 4 }\) = 8 – 1 – 1 = 6

Question 10.

If f (x) = 3x² – 4x, find the value of a given that f ‘(a) = 5.

Solution:

Given f (x) = 3x² – 4x

Diff both sides w.r.t x, we have f'(x) = 6x – 4

∴ f ‘(a) = 6a – 4

Also f ‘(a) = 5

∴ 5 = 6a – 4 ⇒ 6a = 9 ⇒ a = \(\frac { 3 }{ 2 }\)

Question 11.

Differentiate from first principle.

(i) 3x

(ii) (x + 1) (2x – 3)

(iii) \(\frac{2-x}{4+3 x}\)

(iv) x^{-3/4}

(v) \(\sqrt{x}+\frac{1}{\sqrt{x}}(x>0)\).

Solution: