How to find the derivative of $(2x+5)^3(3x-1)^4$ 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: $$ 6(2x+5)^{2}(3x-1)^{3}(7x+9)$$

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.
