# FIND THE DERIVATIVE OF INVERSE TRIGONOMETRIC FUNCTIONS

## The Derivative of an Inverse Trigonometric Function

Here are the formulas to find the derivative of inverse trigonometric functions.

d (sin -1(x)) = 1/√(1 - x2)

d (cos-1(x)) = -1/√(1 - x2)

d (tan-1(x)) = 1/(1 + x2)

d (csc -1(x)) = -1/x√(x2 - 1)

d (sec-1(x)) = 1/x(x2 - 1)

d (cot-1(x)) = -1/(1+ x2)

Problem 1 :

Differentiate y = x tan-1(x)

Solution :

y = x tan-1(x)

Since two functions are multiplied, we have to use the product rule to find the derivative.

Here u = x and v =  tan-1(x)

u' = 1 and v' = 1/(1 + x2)

uv' + vu' = x [1/(1 + x2)] + (1) tan-1(x)

dy/dx = x/(1 + x2) + tan-1(x)

So, the answer is x/(1 + x2) + tan-1(x).

Problem 2 :

Differentiate y = 1/cos-1(x)

Solution :

y = 1/cos-1(x)

Since we have x related function which is at the denominator only, we have to move this to the numerator and find the derivative.

y = [cos-1(x)]-1

We can find the derivative, first using the power rule and we use chain rule.

dy/dx = -1 [cos-1(x)]-2 (-1/√(1 - x2))

= ( -1/ [cos-1(x)]2 ) (-1/√(1 - x2))

dy/dx = 1/[cos-1(x)]2 √(1 - x2)

Problem 3 :

Differentiate y = arc sin x2

Solution :

y = arc sin x2

dy/dx = [1/√(1 - (x2)2)] 2x

dy/dx  = 2x /√(1 - x4)

So, the value of dy/dx is 2x /√(1 - x4)

Problem 4 :

If f(x) = arctan (e-x), then f'(-1) = ?

Solution :

f(x) = arctan (e-x)

d(tan -1x) = 1/(1 + x2)

From the given function,

f'(x) = { 1/(1 + (e-x)2) } (-e-x)

f'(x) = (-e-x)/(1 + (e-2x))

To find f'(1), we have to apply 1 as x in the function f'(x)

f'(1) = (-e-1)/(1 + (e-2(1)))

= (-1/e1)/(1 + (e-2))

= (-1/e1)/(1 + (1/e2))

= (-1/e1)/((e2 + 1) / e2)

= (-1/e1) x {(e2)/(e2 + 1)}

f'(1) = {(-e1)/(1 + e2)}

So, the value of f'(1) is {(-e1)/(1 + e2)}

Problem 5 :

If f(x) = arc tan (sin (x)), then f'(π/3) = ?

Solution :

f(x) = arc tan (sin (x))

f'(x) = 1/(1 + (sin x)2) (cos x)

From the given function,

f'(x) = (1/(1 + (sin x)2)) (cos x)

f'(π/3) = (cos (π/3/ (1 + (sin (π/3) )2)

cos (π/3) = 1/2 and sin (π/3) = √3/2

Applying all these values, we get

f'(π/3) = (1/2) / (1 + (√3/2)2)

= (1/2) / (7/4)

= (1/2) x (4/7)

f'(π/3= 2/7

Problem 6 :

If y = cos(sin -1x) , then y' = ?

Solution :

y = cos(sin -1x)

dy/dx = sin (sin -1x) ( -1/√(1 - x2) )

= (x) ( -1/√(1 - x2) )

dy/dx = ( -x/√(1 - x2) )

So, the value of dy/dx is ( -x/√(1 - x2) )

Problem 7 :

Let f be the function given by f(x) = x tan^-1 (x)

a) Find f'(x)

b) Write an equation for the line tangent to the graph of f at x = 1

Solution :

f(x) = x tan^-1 (x)

a) To find f'(x), we have to use logarithmic differentiation since we have variable at the power.

let y = x tan^-1 (x)

Taking logarithms on both sides,

log y = log (x tan^-1 (x))

log y = tan-1 (x) log (x )

Now finding the derivative using product rule,

u = tan-1 (x) and v = log x

u' = 1/(1 + x2) and v' = 1/x

uv' + vu' =  tan-1 (x) (1/x) + log x (1/(1 + x2))

= (tan-1 (x)/x) +  ( log x /(1 + x2) )

f'(x) =  {((1 + x2)(tan-1 (x)) + x log x)} / x (1 + x2)

b) Equation of tangent line at x = 1

y - y1 = m(x - x1)

f'(x) of x tan^-1 (x) at x = 1

f'(x) =  {((1 + x2)(tan-1 (x)) + x log x)} / x (1 + x2)

At x = 1, f'(x)

f'(x) =  {((1 + 12)(tan-1 (1)) + 1 log (1))} / (1) (1 + 12)

= { 2 (π/4) + 1 (0) } / 1(2)

= (π/4) / 2

Slope at x = 1 is π/8

When x = 1,

Given that f(x) = x tan^-1 (x)

f(1) = x tan^-1 (1)

= x(π/4)

Equation of tangent :

y - x(π/4) = π/8 (x - 1)

Problem 7 :

If f(x) = sin -1(x + 1) , then f'(x) = ?

Solution :

f(x) = sin -1(x + 1)

f'(x) = 1/√(1 - ((x + 1)2)

= 1/√(1 - (x2 + 2x + 1))

= 1/√(1 - x2 - 2x - 1)

f'(x) = 1/√(- x2 - 2x)

So, the answer is 1/√(- x2 - 2x).

Problem 8 :

If f(x) = x2 tan -1(5x) , then f'(x) = ?

Solution :

f(x) = x2 tan -1(5x)

Since two x related functions are multiplying, then to find the derivative we have to use the product rule.

u = x2 and v = tan -1(5x)

u' = 2x and v' = 1/(1 + (5x)2) ==> 1/(1 + 25x2) (5)

Using product rule :

= uv' + vu'

x2 (5/(1 + 25x2)) + tan -1(5x) (2x)

= (5x2/(1 + 25x2)) + 2x tan -1(5x)

So, the answer is (5x2/(1 + 25x2)) + 2x tan -1(5x).

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