Part2.doc

Can someone please help as I have tried this and have gone without sleep for 2 nights.

Thank you.

2. The composite rule is also known as the chain rule. Put simply, the derivative of the inside times the derivative of the outside.

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Where $\displaystyle u=\frac{2}{3}x-\frac{1}{6}x^{2}$

The inside, u, is: $\displaystyle \frac{4x-x^{2}}{6}=\frac{2}{3}x-\frac{1}{6}x^{2}$

Find the derivative of this and get $\displaystyle \frac{du}{dx}=\frac{2}{3}-\frac{1}{3}x$

$\displaystyle \frac{dy}{du}=e^{u}$. And we know the derivative of $\displaystyle e^{u}=e^{u}$, so we have:

$\displaystyle \frac{dy}{dx}=\left(e^{\frac{2}{3}x-\frac{1}{6}x^{2}}\right)\left(\underbrace{\frac{2} {3}-\frac{1}{3}x}_{\text{du/dx}}\right)$

3. In case that was too complicated for you, the following example might be easier. The principle is the same - the derivative of a function is the derivative of the outside times the derivative of the inside.

Suppose you want to take the derivative of $\displaystyle \sin(x^2+3)$

Then your answer is $\displaystyle D(\sin(x^2+3)) * D(x^2+3) = \cos(x^2+3) * 2x$

4. ## composite rule

can you please show me an example with the first question in my attachment as i just can not get my head around it.

Do I keep the e in the fuction.

Basically YES: $\displaystyle \frac{{d\left[ {e^{f(x)} } \right]}}{{dx}} = f'(x)e^{f(x)}$.
$\displaystyle \frac{{d\left[ {e^{x^3 } } \right]}}{{dx}} = \left[ {3x^2 } \right]e^{x^3 }$.
$\displaystyle \frac{{d\left[ {e^{\sin (x)} } \right]}}{{dx}} = \left[ {\cos (x)} \right]e^{\sin (x)}$
$\displaystyle \frac{{d\left[ {e^{\tan ^2 (x)} } \right]}}{{dx}} = \left[ {2\tan (x)\sec ^2 (x)} \right]e^{\tan ^2 (x)}$