How do you take the derivative of #tan^-1 2x#?

1 Answer
Aug 4, 2015

Use the derivative of #tan^-1# and the chain rule.

Explanation:

The derivative of #tan^-1x# is #1/(1+x^2)# (for "why", see note below)

So, applying the chain rule, we get:

#d/dx(tan^-1u) = 1/(1+u^2)*(du)/dx#

In this question #u = 2x#, so we get:

#d/dx(tan^-1 2x) = 1/(1+(2x)^2)*d/dx(2x)#

# = 2/(1+4x^2)#

Note

If #y = tan^-1x#, then #tany = x#

Differentiating implicitly gets us:

#sec^2y dy/dx = 1," "# so

#dy/dx = 1/sec^2y#

From trigonometry, we know that #1+tan^2y = sec^2y#

so #dy/dx = 1/(1+tan^2y)#

and we have #tany = x#, so we get:

For #y = tan^-1x#, the derivative is:

#dy/dx = 1/(1+x^2)#