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QUESTION: 1

If y = tan^{-1} x, then in terms of y

Solution:

y = tan^{-1} x

=> tan y = tan(tan^{-1} x)

x = tan y

sec^{2} y dy/dx = 1

=> 1/(cos^{2}y) dy/dx = 1

dy/dx = cos^{2}y

Differentiate it with respect to x

d^{2}y/dx^{2} = 2cos y(-sin y)(dy/dx)..................(1)

Put the value of dy/dx in eq(1)

=> d^{2}y/dx^{2} = 2cos y(-sin y)(cos^{2} y)

=> d^{2}y/dx^{2} = 2cos^{3} y siny

QUESTION: 2

Solution:

x^{a} y^{b} = (x−y)^{a}+b

taking log both the sides.

log(x^{a} y^{b})=log(x−y)^{(a+b)}

alogx+blogy=(a+b)log(x−y)

differentiating both sides w.r.t 'x'.

(a/x+b/y)dy/dx = (a+b)/(x−y)[1−dy/dx]

dy/dx[b/y+(a+b)/(x−y)] = (a+b)/(x−y) − a/x

dy/dx[(bx−by+ay+by)/y(x−y)] = (ax+bx−ax+ay)/x(x−y)

dy/dx[(bx+ay)/y]=(bx+ay)/x

dy/dx = y/x

QUESTION: 3

If f(x) = x + cot x,

Solution:

f(x) = x + cot x

f’(x) = 1 + (-cosec^{2} x)

f”(x) = 0 - 2cosec x(-cosec x cot x)

= 2 cosec^{2} x cot x

f”(π/4) = 2 cosec^{2} (π/4) cot(π/4)

= 2 [(2)^½]^{2} (1)

= 4

QUESTION: 4

Find the differential coefficient y = (sec 5x)^{5x}

Solution:

QUESTION: 5

Differentiate

Solution:

QUESTION: 6

If e^{y}(x+5) = 1, then evaluate

Solution:

e^{y} =1/ (x+5)

Taking log both side we get

log e^{y} = log {1/(5+x)}

then, y = log {1/(5+x)}

Differentiating both side ,

dy/dx =( x+5) . {-1/(5+x)²}

dy/dx = -1/(5+x) ……..( 1)

Again Differentiating, d²y/dx²= 1/(5+x)²

d²y/dx²= {-1/(5+x)}²

From equation (1)

d²y/dx² = {dy/dx}²

QUESTION: 7

Find from x = 2at^{2}, y = at^{4}

Solution:

y = at^{4}, x = 2at^{2}

dy/dt = 4at^{3} dx/dt = 4at => dt/dx = 1/4at

Divide dy/dt by dx/dt, we get

dy/dx = t^{2}

d^{2} y/dx^{2} = 2t dt/dx……………….(1)

Put the value of dt/dx in eq(1)

d^{2} y /dx^{2} = 2t(1/4at)

= 1/2a

QUESTION: 8

The differential coefficient of tan^{-1} (log x) with respect to x is :

Solution:

QUESTION: 9

Solution:

QUESTION: 10

Find the second derivative of e^{x}cosx

Solution:

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