Does the chain rule apply in inverse function derivatives?

Question

My problem is finding the derivative of $y=\arctan (3x)$. Would it be

$$y'= \dfrac{1}{1+(3x)^2}$$

or

$$y'= \dfrac{1}{1+(3x)^2}\times 3$$

Answer 1

Umm...Second one because

$$y=f(g(x))$$ $$y'=f'(g(x)) \cdot g'(x)$$

Now put $f(x)=\arctan(x)$ and $g(x)=3x$ $$y=\arctan(g(x))$$

$$y'=\frac{1}{1+(g(x))^2}\cdot g'(x)=\frac{3}{1+(3x)^2}$$

This was added as a response to OP's comment

$$\frac{d}{dx}\arcsin(e^{2x})=\frac{1}{\sqrt{1-(e^{2x})^2}}\cdot e^{2x}\cdot2=\frac{2e^{2x}}{\sqrt{1-e^{4x}}} $$

Answer 2

Yes, the chain rule applies to derivatives of inverse trigonometric functions.

$$y'= \dfrac{1}{1+(3x)^2}\times 3$$