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:

- Find dy/dx for (9x-3x^2)^3 [make this part A]
- find dy/dx for cos(4x^3 - 3x) [make this part B]
- find dy/dx for e^(7 - 12x) [make this part C]
- find dy/dx for ln(-x^2 + 3) [make this part D]
- product rule A and B [make this part E]
- product rule C and D [make this part F]
- quotient rule E and F

Does this look right?

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

Quote:

Originally Posted by

**HallsofIvy** It is simply $\displaystyle \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{fg}$ 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: $\displaystyle \left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}$ ?

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

You are correct.

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