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

• May 23rd 2011, 06:57 PM
ezzab69
Finding the derivative of a large function. Help with the order of operations.
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:
1. Find dy/dx for (9x-3x^2)^3 [make this part A]
2. find dy/dx for cos(4x^3 - 3x) [make this part B]
3. find dy/dx for e^(7 - 12x) [make this part C]
4. find dy/dx for ln(-x^2 + 3) [make this part D]
5. product rule A and B [make this part E]
6. product rule C and D [make this part F]
7. quotient rule E and F

Does this look right?
• May 23rd 2011, 07:04 PM
skeeter
$\frac{d}{dx} \left(\frac{uv}{wz}\right) = \frac{(wz)(uv'+vu')-(uv)(wz'+zw')}{(wz)^2}$
• May 23rd 2011, 07:07 PM
ezzab69
Thank you so much. I have never seen this version of the quotient rule before.
• May 24th 2011, 03:06 AM
Ackbeet
As an alternative, you could use logarithmic differentiation.
• May 24th 2011, 05:05 AM
HallsofIvy
Quote:

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'
• Oct 23rd 2011, 07:44 PM
ezzab69
Re: Finding the derivative of a large function. Help with the order of operations.
Quote:

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: $\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$ ?
• Nov 4th 2011, 09:50 AM
sbhatnagar
Re: Finding the derivative of a large function. Help with the order of operations.
Quote:

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