FIND THE DERIVATIVE OF INVERSE TRIGONOMETRIC FUNCTIONS

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).