# Derivatives

• Nov 17th 2012, 03:13 PM
xRockbottomx
Derivatives
I am having some issues with Derivatives. My first question is what is the derivative of e^4x? I know one of the rules is (e^x)' = e^x. Then why is (e^4x)' not e^4x?

Thanks
• Nov 17th 2012, 03:16 PM
Re: Derivatives
You have to apply what is known as the Chain Rule in the case of e^4x. The Chain rule is f(g(x))' = f'(g(x)) * g'(x). f'(x) = e^x , g(x) = 4x, g'(x) = 4

(e^4x)' = e^4x * 4 = 4e^4x
• Nov 17th 2012, 03:28 PM
xRockbottomx
Re: Derivatives
When do you use the chain rule?
• Nov 17th 2012, 03:29 PM
xRockbottomx
Re: Derivatives
and thank you
• Nov 17th 2012, 04:16 PM
Another (easier) way to write the chain rule is \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \end{align*}. So you let your "inner" function of x be u, so that you can then write your function y in terms of u, take the derivative of each, then convert \displaystyle \begin{align*} \frac{dy}{du} \end{align*} back to a multiple of x and multiply them.
For example, if our function was \displaystyle \begin{align*} y = (x + 3)^2 \end{align*}, our "inner" function is \displaystyle \begin{align*} u = x + 3 \end{align*} leaving \displaystyle \begin{align*} y = u^2 \end{align*}.
Differentiating each gives \displaystyle \begin{align*} \frac{du}{dx} = 1 \end{align*} and \displaystyle \begin{align*} \frac{dy}{du} = 2u = 2(x + 3) \end{align*}.
Therefore the derivative is \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2(x + 3) \cdot 1 = 2(x + 3) \end{align*}.