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How do you find #d/dx(arctan(e^2x))#?

Answer 1

First of all, I am assuming that there is a mistake in the formatting, and that you mean #e^{2x}# instead of #e^2x#.

This is a composite function. Which means, a function #f# of the form #f(x)=g(h(x))#. In your case, #g(x)=\arctan(x)#, and #h(x)=e^{2x}#. To derive such functions, you must use the chain rule, which states that, using the same notation I introduced above, #f'(x)=g'(h(x))*h'(x)#. The two derivatives we need are both elementary, so I will not proof the results: #d/{dx} \arctan(x)=1/{x^2+1}# #d/dx e^{2x}=2e^{2x}#

Note that for the second derivative, we actually use the chain rule one more time, because #e^{2x}# is of the form #a(b(x)#, with #a(x)=e^x#, and #b(x)=2x#.

The final answer is thus, as said above, #g'(h(x))*h'(x)=1/{e^{4x}+1}*2e^{2x}={2e^{2x}}/{e^{4x}+1}#

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Answer 2

Christopher Agresta

To find ( \frac{d}{dx}(\arctan(e^{2x})) ), you can use the chain rule. The derivative is ( \frac{2e^{2x}}{1 + e^{4x}} ).

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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.