How to find a derivative of the following function?
$$\ f(x)=(2x+5)^{3} (3x-1)^{4}$$ So I used: $$(fg)'= f'g + fg'$$ and $$(f(g(x)))'= f'(g(x)) + g'(x)$$
Then I got:
$$ f(x)= 6(2x+5)^{2} + 12(3x-1)^{3}$$
and I don't know how to get the solution from this, which is: $$ 6(2x+5)^{2}(3x-1)^{3}(7x+9)$$
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$\begingroup$The first half of this answer addresses an earlier (apparently not intended) question from the OP, and the question asked has since been edited.
You have a difference of functions, not a product of functions:
$$h(x) = [f(x)]^3 - [g(x)]^3$$. So you need to use the chain rule and power rule here on each , in each term of the difference, not the product rule:
$$h(x) = 3[f(x)]^2\cdot f'(x) = 3[g(x)]^2 \cdot g'(x)$$ $$ $$ $$\begin{align} \ h(x)=(2x+5)^{3} -(3x-1)^{3} \implies h'(x) & = 3(2x + 5)^2(2x + 5)' - 3(3x-1)^2(3x-1)'\\ \\ & = 3(2x + 5)^2(2) - 3(3x - 1)^2(3)\\ \\ & = 6(2x + 5)^2 - 9(3x-1)^2\end{align}$$
Edited question and new response
Now you've got the product rule down right!
Using the product rule, we get $$f'(x)=(2x+5)^3\underbrace{\Big((3x-1)^4\Big)'}_{\text{apply chain rule}} +\underbrace{\Big((2x+5)^3\Big)'}_{\text{apply chain rule}}(3x-1)^4.\tag{1}$$
Your primary mistake here was not understanding, or correctly applying, the chain rule.
$$f(x) = h(g(x))\implies f'(x) = h'(g(x))\cdot g'(x)$$ Here, applied to $(1)$ to finish off the derivative, gives us $$\Big((3x - 1)^4\Big)' = 4(3x-1)^3\cdot (3x - 1)' = 12(3x-1)^3$$ $$\Big((2x+5)^3\Big)' = 3(2x+5)^2\cdot (2x + 5)' = 6(2x+5)^2$$
Now simply put it all together!
Feel free to check in and comment below if you'd like to verify what you end with.
$\endgroup$ 3 $\begingroup$I assume that you are supposed to have $$f(x)=(2x+5)^3(3x-1)^4,$$ instead of what you've written. By product rule, we have $$f'(x)=(2x+5)^3\bigl[(3x-1)^4\bigr]'+\bigl[(2x+5)^3\bigr]'(3x-1)^4.$$ Applying chain rule (which you've incorrectly stated, by the way) gives us $$f'(x)=(2x+5)^3\cdot 4(3x-1)\cdot(3x-1)'+(3x-1)^4\cdot3(2x+5)^2\cdot(2x+5)',$$ or $$f'(x)=12(2x+5)^3(3x-1)^3+6(2x+5)^2(3x-1)^4.$$ Now, pull out the common factor $6(2x+5)^2(3x-1)^3,$ and see what happens.
$\endgroup$ 1 $\begingroup$$\left(f\left(g\left(x\right)\right)\right)'=f'\left(g\left(x\right)\right)g'\left(x\right)$ and not $f'\left(g\left(x\right)\right)+g'\left(x\right)$
(comment doesn't work for some reason so I post it as answer)
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