**Example 1 :**

Differentiate sin^{-1} (3x - 4x^{3}) with respect to x

**Solution :**

Let y = sin^{-1} (3x - 4x^{3})

First let us consider (3x - 4x^{3}).

If we plug sin θ instead of x in the given function we will get 3 sin θ - 4 sin³ θ.

This the formula for sin 3θ.

So, put x = sin θ and θ = sin^{-1} x

y = sin^{-1} [ 3(sin θ) - 4(sin θ)³ ]

y = sin^{-1} [ 3 Sin θ - 4 sin³ θ ]

y = sin^{-1} [ Sin 3θ ]

y = 3 θ

y = 3 sin^{-1}x

differentiating with respect to x on both sides

dy/dx = 3(1/√(1 - x^{2}))

dy/dx = 3/√(1 - x^{2})

**Example 2 :**

Differentiate tan^{-1} [(1+x^{2})/(1-x^{2})] with respect to x

**Solution : **

Take y = tan^{-1}[(1+x^{2})/(1-x^{2})]

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

So the function has become y = tan^{-1} t

To differentiate this function with respect to x we have to write the formula required.

dy/dx = (dy/du) x (dt/dx)

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

For differentiating this we have to apply the quotient rule. So u = 1+x^{2} and v = 1-x^{2}

u' = 0 + 2x v' = 0 - 2x

u' = 2x v' = -2x

**(U/V)' = [VU' - UV'] /V²**

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

dt/dx = [(2x - 2x^{3}) - (-2x - 2x^{3})]/(1-x^{2})²

dt/dx = [(2x - 2x^{3} + 2x + 2x^{3})]/(1-x^{2})²

dt/dx = [4x/(1-x^{2})²]

y = tan^{-1} t

dy/dt = 1/(1+t²)

dy/dt = 1/(1+[(1+x^{2})/(1-x^{2})]²)

dy/dt = 1/(1+[(1+x^{2})²/(1-x^{2})²])

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

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

dy/dt = (1-x^{2})]^{2}/[(1 + x^{4} - 2x^{2})+(1+ x^{4} + 2x^{2})]

dy/dt = (1-x^{2})²/(2 + 2x^{4})

dy/dx = (dy/dt) **x** (dt/dx)

= (1-x^{2})²/(2 + 2x^{4}) **x** 4x/(1-x^{2})^{2}

= 4x/(2 + 2x^{4})

= 4x/2(1+x^{4})

dy/dx = 2x/(1+x^{4})

**Example 3 :**

Differentiate the following

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

**Solution :**

Let y = sin^{-1}[2x/(1+x^{2})]

Let x = tan a

a = tan^{-1}x

y = sin^{-1}[2tan a/(1+tan^{2} a)]

y = sin^{-1}[sin2a]

y = 2tan^{-1}x

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

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

**Example 4 :**

Differentiate the following

tan^{-1}[√(1 - cos x)/(1+cosx)]

**Solution :**

Let y = tan^{-1}[√(1 - cos x)/(1+cosx)]

Trigonometric formula for (1-cos x)/(1+cosx) = tan^{2}(x/2)

By applying trigonometric formula, we may find derivatives easily.

y = tan-1[√tan^{2}(x/2)]

y = x/2

Differentiating with respect to "x"

dy/dx = 1/2

**Example 5 :**

Differentiate the following

tan^{-1}[6x/1-9x^{2}]

**Solution :**

Let y = tan^{-1}[6x/1-9x^{2}]

y = tan^{-1}[2(3x) /1-(3x)^{2}]

Let 3x = tan θ

y = tan^{-1}[2 tan θ /1-tan^{2}θ]

2 tan θ /1-tan^{2}θ = tan 2θ

y = tan^{-1}[tan 2θ]

y = 2θ

dy/dx = 2 (dθ/dx) ------(1)

3x = tan θ

3(1) = sec^{2}θ(dθ/dx)

3/sec^{2}θ = (dθ/dx)

Applying dθ/dx = 3/sec^{2}θ in (1)

dy/dx = 2 (3/sec^{2}θ)

= 2(3)/(1 + tan^{2}θ)

= 6/(1 + (3x)^{2})

= 6/(1 + 9x^{2})

**Example 6 :**

Differentiate the following

cos (2tan^{-1}[√(1-x)/(1+x)])

**Solution :**

Let y = cos (2tan^{-1}[√(1-x)/(1+x)])

let x = cos θ

Differentiating with respect to x, we get

dx/dθ = -sin θ

y = cos (2tan^{-1}[√(1-cosθ)/(1+cosθ)])

y = cos (2tan^{-1}[√tan^{2}(θ/2])

y = cos (2tan^{-1}[tan(θ/2])

y = cos θ

Differentiating with respect to θ.

dy/dx = -sin θ (dθ/dx) -----(1)

Now let us apply the value of dx/dθ = -sin θ in (1)

dy/dx = -sin θ (-1/sinθ)

dy/dx = 1

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