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

1. ## 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?

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

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

4. As an alternative, you could use logarithmic differentiation.

5. 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'

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

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

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

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