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

Answer 1

Ezra Adams

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

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)#

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Answer 2

William Abdalla

To take the derivative of ( \tan^{-1}(2x) ), use the chain rule, which states that the derivative of an outer function applied to an inner function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. The derivative of ( \tan^{-1}(u) ) is ( \frac{1}{1+u^2} ), and the derivative of ( 2x ) with respect to ( x ) is ( 2 ). Therefore, the derivative of ( \tan^{-1}(2x) ) is ( \frac{2}{1+(2x)^2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.