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QUESTION: 1

Derivative of cos with respect to x is

Solution:

QUESTION: 2

Differentiate with respect to x.

Solution:

(cosx − sinx)/(cosx + sinx) = (1 − tanx)/(1 + tanx)

tan(A − B) = (tanA − tanB)/(1 + tanAtanB)

= tan(π/4−x)

putting this value in question.

tan^{−1} tan(π/4−x)

π/4 − x.

so d(π/4 − x)/dx = -1

QUESTION: 3

The dervative id sec^{-1} x is

Solution:

y = sec^{−1} x

by rewriting in terms of secant,

⇒ secy = x

by differentiating with respect to x,

⇒ secy tany y'= 1 by dividing by secy tany,

⇒ y'= 1/secy tany

since secy = x and tanx = √sec^{2}y − 1 = √x^{2}−1

⇒ y'= 1/x√x^{2}−1

Hence, d/dx(sec^{−1}x) = 1/x√x^{2}−1

QUESTION: 4

If y + sin y = 5x, then the value of dy/dx is

Solution:

y + sin y = 5x

dy/dx + cos ydy/dx = 5

dy/dx = 5/(1+cos y)

QUESTION: 5

Derivative of tan^{-1} x is

Solution:

QUESTION: 6

If xy = 2, then dy/dx is

Solution:

xy = 2

x dy/dx + y = 0

x dy/dx = -y

dy/dx = -y/x

QUESTION: 7

If 3 sin(xy) + 4 cos (xy) = 5, then = .....

Solution:

3sinxy + 4cosxy = 5

⇒ 5(3/5 sinxy + 4/5 cosxy) = 5

⇒ (3/5 sinxy + 4/5 cosxy) = 1

now (3/5)²+(4/5)² = 1

so let, 3/5 = cosA

⇒ 4/5 = sinA

So , (3/5 sinxy + 4/5 cosxy) = 1

⇒ (cosAsinxy + sinAcosxy) = 1

⇒ sin(A+xy) = 1

⇒ A + xy = 2πk + π/2 (k is any integer)

⇒ sin⁻¹(4/5) + xy = 2πk + π/2

differenciating both sides with respect to x

0 + xdy/dx + y = 0

dy/dx = -y/x

QUESTION: 8

Solution:

y = tan^{-1}(1-cosx)/sinx

y = tan^{-1}{2sin^{2}(x/2)/(2sin(x/2)cos(x/2)}

y = tan^{-1}{tan x/2}

y = x/2 => dy/dx = 1/2

QUESTION: 9

Solution:

cos^{-1}[(1-x^{2})/(1+x^{2})]

Put x = tanθ

y = cos^{-1}(1 - tan^{2}θ)/(1 + tan^{2}θ)

y = cos^{-1}(cos^{2}θ)

y = 2θ……………….(1)

Put θ = tan^{-1} x in eq(1)

y = 2(tan^{-1} x)

dy/dx = 2(1/(1+x^{2}))

dy/dx = 2/(1+x^{2})

QUESTION: 10

Solution:

now differentiate y with respect to x,

dy/dx = -{[x d/dx(1+x) - (1+x)dx/dx]}/(1+x^{2})

= -1/(1+x)^{2}

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