Finding the derivative of a large function. Help with the order of operations.

**ezzab69** · May 23rd 2011, 05:57 PM

Differentiate

(9x -3x^2)^3 cos(4x^3 – 3x)
e^(7-12x) ln(-x^2 + 3)

With respect to x

I don't want the solution, what i'm after is some help in how to tackle this differentiation question.

This is what i plan to do:

Does this look right?

**skeeter** · May 23rd 2011, 06:04 PM

$\frac{d}{dx} \left(\frac{uv}{wz}\right) = \frac{(wz)(uv'+vu')-(uv)(wz'+zw')}{(wz)^2}$

**ezzab69** · May 23rd 2011, 06:07 PM

Thank you so much. I have never seen this version of the quotient rule before.

**Ackbeet** · May 24th 2011, 02:06 AM

As an alternative, you could use logarithmic differentiation.

**HallsofIvy** · May 24th 2011, 04:05 AM

Originally Posted by ezzab69

Thank you so much. I have never seen this version of the quotient rule before.

It is simply $\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{fg}$ with f= uv and g= wz. Of course, f'= (uv)'= u'v+ uv' and g'= (wz)'= w'z+ wz'

**ezzab69** · October 23rd 2011, 06:44 PM

Originally Posted by HallsofIvy

It is simply $\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{fg}$ with f= uv and g= wz. Of course, f'= (uv)'= u'v+ uv' and g'= (wz)'= w'z+ wz'

Just quickly clarify, you're using the quotient rule right? So shouldn't it be: $\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$ ?

**sbhatnagar** · November 4th 2011, 08:50 AM

Originally Posted by ezzab69

Just quickly clarify, you're using the quotient rule right? So shouldn't it be: $\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$ ?

You are correct.
$\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$

Thread: Finding the derivative of a large function. Help with the order of operations.