• Jun 6th 2008, 04:51 PM
omkara
Attachment 6667

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

Thank you.
• Jun 6th 2008, 05:25 PM
galactus
The composite rule is also known as the chain rule. Put simply, the derivative of the inside times the derivative of the outside.

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

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

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

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

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

$\frac{dy}{dx}=\left(e^{\frac{2}{3}x-\frac{1}{6}x^{2}}\right)\left(\underbrace{\frac{2} {3}-\frac{1}{3}x}_{\text{du/dx}}\right)$
• Jun 6th 2008, 09:06 PM
vxcv
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 $\sin(x^2+3)$

Then your answer is $D(\sin(x^2+3)) * D(x^2+3) = \cos(x^2+3) * 2x$
• Jun 7th 2008, 09:38 AM
omkara
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.

• Jun 7th 2008, 12:12 PM
Plato
Quote:

Originally Posted by omkara
Do I keep the e in the fuction.

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

$\frac{{d\left[ {e^{\sin (x)} } \right]}}{{dx}} = \left[ {\cos (x)} \right]e^{\sin (x)}$

$\frac{{d\left[ {e^{\tan ^2 (x)} } \right]}}{{dx}} = \left[ {2\tan (x)\sec ^2 (x)} \right]e^{\tan ^2 (x)}$