Differentiation Class 12 OP Malhotra Exe-8E ISC Maths Solutions Ch-8 Solutions. In this article you would learn about differentiation of e^x  and a^x . Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

 Differentiation Class 12 OP Malhotra Exe-8E ISC Maths Solutions Ch-8

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-8	Differentiation
Writer	OP Malhotra
Exe-8(E)	differentiation of ex  and ax

Exercise- 8E

Differentiation Class 12 OP Malhotra Exe-8E Solution

Que-1: (i) e^3x
(ii) e^{cos x}
(iii) e^-x/2
(iv)e^x²+2x
(v) e^(√1+x+x²)
(vi) e^{√sin x}
(vii) e^(x²/1+x²)

Sol: (i) Let y = e^3x
∴ dy/dx=(e^3x)d/dx (3x) = [∵ d/dx e^x = e^x]
= 3e^3x

(ii) Let y = e^{cos x};
Diff both sides w.r.t. x ; we get
dy/dx=(e^{cos x}) d/dx (cos x)  = – sin x e^{cos x}

(iii) Let y = e^-x/2
Diff both sides w.r.t. x ; we get
∴ dy/dx=(e^-x/2)d/dx (-x/2) =-1/2 e^-x/2

(iv) Let y = e^x²+2x ;
Diff both sides w.r.t. x ; we get
∴ dy/dx=e^x²+2x d/dx (x²+2x)
= 2(x + 1)e^x²+2x

(v) Let y =e^(√1+x+x²)
Diff both sides w.r.t. x ; we get
∴ dy/dx=e^(√1+x+x²) d/dx (√1+x+x²)
= e^(√1+x+x²) 1/2 (1+x+x²)^(1/2-1) d/dx (1+x+x²)
= e^(√1+x+x²) 1/2√(1+x+x²) (1+2x)

(vi) Let y= e^{sin √x}

(vii) Let y=e^(x²/1+x²)

Que-2: (i) 3^x
(ii) 8^{cos x}
(iii) a^{sin x}
(iv) a^3x²
(v) 5^{log sin x}
(vi) 10^10x
(vii) 2^(x/log x)

Sol: (i) Let y = 3^x
Diff both sides w.r.t. x ; we get
dy/dx=3^x log 3 [∵ d/dx = a^x log a]

(ii) Let y = 8^{cos x}
Diff both sides w.r.t. x ; we get
dy/dx=(8^{cos x})log 8 d/dx (cos x)
= – sin x 8^{cos x} log 8

(iii) Let y = a^{sin x}
Diff both sides w.r.t. x ; we get
dy/dx=(a^{sin x}) log a d/dx (sin x)
= a^{sin x} log a. cos x

(iv) Let y = a^3x²
Diff both sides w.r.t. x ; we get
dy/dx=(a^3x²) log a . 6x

(v) Let y = 5^{log sinx}

Que-3: (i) x e^-x
(ii) e^x cot x
(iii) e^ax bx
(iv) (e^x)/x
(v) (e^x)/(1+sinx)
(vi) x e^x²
(vii)e^{x sinx}
(viii) e^ax cos (bx + c)
(ix) x² e^x sin x
(x) sin (e^x log x)
(xi) e^ax cos (b tan x)
(xii) (e^x²) log₁₀ (2x)

Sol: (i) Let y = x e^-x
Diff both sides w.r.t. x ; we get
dy/dx= x d/dx (e^-x) + (e^-x) d/dx (x)
= x e^-x (-1) + e^-x.1 = (1 – x)e^-x

(ii) Let y = e^x cot x
Diff both sides w.r.t. x ; we get
dy/dx= (e^x) d/dx (cot x) + cot x d/dx (e^x)
= e^x (-cosec²x) + cot x e^x
= e^x [cot x – cosec²x]

(iii) Let y = e^ax sin bx
Diff both sides w.r.t. x ; we get
dy/dx=(e^ax) d/dx (sin bx) + sin bx d/dx (e^ax)
= b e^ax cos bx + a sin bx e^ax = em (b cos bx + a sin bx)

(iv) Let y =(e^x)/x;
Diff both sides w.r.t. x ; we get

(v) Let y = (e^x)/(1+ sin x)
Diff both sides w.r.t. x ; we get

(vi) Let y = x e^x² ;
Diff both sides w.r.t. x ; we get
dy/dx= x d/dx (e^x²) + (e^x²) d/dx (x) = x . (e^x²) . 2x + (e^x²) . 1 = (e^x²)(2x²+1 )

(vii) Let y = e^x sin x
Diff both sides w.r.t. x ; we get
dy/dx=e^x sin x d/dx (x sinx) = e^x sin x[x. cos x + sin x]
= e^{x sinx} (xcosx + sinx)

(viii) Let y = e^ax cos (bx + c)
Diff both sides w.r.t. x ; we get
dy/dx=(e^ax) d/dx cos (bx + c) + cos (bx + c) d/dx (e^ax)
= e^ax {-sin(bx + c)}b + cos(bx + c)e^ax.a
= e^ax [a cos(bx + c) – b sin(bx + c)]

(ix) Let y = x² e^x sin x
Diff both sides w.r.t. x ; we get

(x) Let y = sin (e^x log x) ;
Diff both sides w.r.t. x ; we get
dy/dx= cos(e^x log x)d/dx (e^x log x)
= cos(e^x log x)[(e^x).1/x + log x . e^x]
= (e^x)cos(e^x log x)[1/x + log x]

(xi) Let y = e^ax cos (b tan x)
Diff both sides w.r.t. x ; we get

(xii) Let y = e^x log₁₀(2x);
Diff both sides w.r.t. x ; we get

Que-4: (i) log (e^x + e^-x)
(ii) log{e^x/(e^x+1)}
(iii)(e^x)-(e^-x)/(e^x)+(e^-x)
(iv) log x+ e^√x
(v) (e^tanx ).log tan x
(vi) (e^x)+ log x / sin 3x

Sol: (i) Let y = log (e^x + e^-x) ;
Diff both sides w.r.t. x ; we get
dy/dx= 1/(e^x + e^-x) d/dx (e^x + e^-x)
=(e^x – e^-x)/(e^x + e^-x)

(ii) Let y = log{e^x/(e^x+1)}
= log e^x – log (e^x + 1)
⇒ y = x log e – log (e^x + 1)
⇒ y = x – log (e^x + 1) ;
Diff both sides w.r.t. x ; we get
dy/dx= 1- 1/(e^x + 1)e^x
=e^x + 1 – e^x /e^x + 1 = 1/e^x + 1

(iii) Let y = (e^x – e^-x)/(e^x + e^-x)
Diff both sides w.r.t. x ; we get

Que-5: e^{log(x+√(x²-a²))}

Sol:

Que-6: e^{-ax² sin(log x)}

Sol:

Que-7: log x . e^{(tan x + x²)}

Sol:

Que-8: e^x log sin 2x

Sol:
Let y = e^x log sin 2x ;
Diff both sides w.r.t. x ; we get
dy/dx= (e^x )1/sin 2x d/dx (sin 2x) + log sin2x d/dx (e^x)
(e^x )1/sin 2x (cos 2x) . 2 + log sin 2x . (e^x ) = (e^x )( 2cot 2x + log sin 2x)

Que-9: (e^sinx).sin e^x

Sol:

Que-10: e^x log(1 + x²)

Sol:

–: End of Differentiation Class 12 OP Malhotra Exe-8E ISC Math Ch-8 Solution :–

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