I've been having some difficulities with this but I think I figured it out. Could you double check me and make sure I'm doing the correct steps?
Differentiate and simplify the answer.
f(x) = 2(3x+1)^4 (5x-3)^2
=2(3x+1)^4 [2(5x-3)(5)] + [8(3x+1)^3] (5x-3)^2
=2(3x+1)^4 [10(5x-3)] + [24(3x+1)^3] (5x-3)^2
=2(3x+1)^4(5x-3)[34(3x+1)^3(5x-3)]
I hope I'm getting this now.
On the very first step, you forgot to use the chain rule in the second half of the sentence.
Looks like you have the concept of deriving the product of two equations correct.
Becky, here's a suggestion for doing the chain rule. Construct a table like this: Let g = (3x+1) and h = (5x-3).
Then g' = 3 and h' = 5.
What we have then is:
f = 2(g^4)(h^2)
Now differentiate the symbols first, then substitute for them.(We treat f,g,h as functions of x.)
So, f' = 2[(4g^3g')h^2 + g^4(2hh')]
Now factor out 2(g^3)h to get,
f' = 2*2*(g^3)h[2g'h + gh']
Now substitute for the symbols above. Further simplifying might be possible too.
If you use this approach you should make fewer errors I think. In first yr calc you'll probably see longer expressions than this one, some with 3 or even 4 functions and composite functions. Creating a table of symbols and doing the calc on them will save a lot of writing and potential errors.
It seems like you're on the right track, but there are a few mistakes in your work. Let me walk you through the correct steps for differentiating and simplifying the given equation.
Given: f(x) = 2(3x+1)^4 (5x-3)^2
Step 1: Apply the product rule.
The product rule states that if you have two functions, u and v, then the derivative of their product is given by (u'v + uv').
Let's differentiate u = (3x+1)^4 and v = (5x-3)^2 separately.
For u:
u' = 4(3x+1)^3 * (d/dx) (3x+1)
= 4(3x+1)^3 * 3
For v:
v' = 2(5x-3)^1 * (d/dx) (5x-3)
= 2(5x-3)^1 * 5
Step 2: Plug in the derivatives into the product rule formula.
f' = u'v + uv'
= [4(3x+1)^3 * 3] * [2(5x-3)^1] + [2(3x+1)^4] * [2(5x-3)^1 * 5]
Simplifying this expression gives you the derivative of f(x), which is f'(x).