Differentiation inverse trigonometric functions

Question

Find the derivative of

 

tan^-1(1/t)

 

I have seen other answers online but they are really hard to follow. I thought you would need to use the product rule. Not to sure though. I know the derivative of tan^-1 = 1/(1+x²)

Answer 1

As you point out, d(tan^-1t)/dt = 1/(1+t²).  But here, you have tan-1(1/t).  So you need to use the chain rule (not the product rule):

 

      df(u)/dt = df(u)/du * du/dt

 

Let u = 1/t:

 

d(tan^-1u)/dt = d(tan^-1u)/du * du/dt

 

     = 1/(1+u²) * d(t^-1)/dt

 

     = 1/(1+(1/t)²) * (-1/t²)

 

     = -1/(t²+1)

Answer 2

Rewrite the function so that the tangent argument has an exponent.  This makes the derivative process easier to do and we can avoid using the quotient rule.

 

tan^-1(t^-1)

 

 

Now just use the chain rule to derive.  The derivative of the inside function multiplied by the derivative of the outer function evaluated inside.

 

-t^-2 / (1 + (t^-1)²) =

 

-t^-2 / (1 + t^-2) =

 

-(1 / t²) / [1 + (1 / t²)] =

 

-(1 / t²) / [(t² + 1) / t²] =

 

- 1 / (t² + 1)

Answer 3

This is the chain rule.....like y=sin (2x) leads to cos ( 2x) *2 because of the chain rule.

 

The derivative of tan^-1 u = 1/(1+u²).

 

The derivative of tan^-1 (1/t)  is therefore 1/(1+(1/t)²) * (-t^-2).

 

This simplified is -1/(t²+1).

 

Hope this helped