Derivatives

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November 17th 2012, 02:13 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
Adam
November 17th 2012, 02:16 PM
MacstersUndead

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
November 17th 2012, 02:28 PM
xRockbottomx

Re: Derivatives

When do you use the chain rule?
November 17th 2012, 02:29 PM
xRockbottomx

Re: Derivatives

and thank you
November 17th 2012, 03:16 PM
MacstersUndead

Re: Derivatives

You use the chain rule whenever you are faced with a form like f(g(x)) and you know the derivative of f(x) and g(x)
November 17th 2012, 04:41 PM
Prove It

Re: Derivatives

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 $\displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2(x + 3) \cdot 1 = 2(x + 3) \end{align*}$ .