Solution assignment 01 Chain rule

Differentiate:

$y=(3x+4)^3$

We begin with the formal method. We take:

$u=3x+4$

and get:

$y=u^3$

$\displaystyle\frac{dy}{du}=3u^2$

$\displaystyle\frac{du}{dx}=3$

We apply the chain rule:

$\displaystyle\frac{dy}{dx}=\displaystyle\frac{dy}{du}\displaystyle\frac{du}{dx}=3u^2\cdot3=9(3x+4)^2$

According to the informal method we get:

$y'=3(3x+4)^2$ times the derivative of the function which is on the place of the original $x$ in the standard function.

Thus:

$y'=3(3x+4)^2\cdot3=9(3x+4)^2$