Differentiate y= (2x-5)^4 (8x^2-5)^-3

I've gotten as far as
-3(2x-5)^4 (8x^2-5)^-4 (16x) + 4(8x^2-5)^-3(2x-5) * 2

The answer is 8(2x-5)^3(8x^2-5)^-4(-4x^2+30x-5)

How do I get the answer?

Question

Differentiate y= (2x-5)^4 (8x^2-5)^-3

I've gotten as far as 
-3(2x-5)^4 (8x^2-5)^-4 (16x) + 4(8x^2-5)^-3(2x-5) * 2

The answer is 8(2x-5)^3(8x^2-5)^-4(-4x^2+30x-5)

How do I get the answer?

amistre64 · Answer

product rule, base rule, chain rule

amistre64 · Answer

totally saw that wrong

amistre64 · Answer

power rules not exponent rules

amistre64 · Answer

$$y = f^p~g^q$$
$$y' = pf^{p-1}~f'~g^q+f^p~q~g^{q-1}~g'$$

amistre64 · Answer

$$ y= (2x-5)^4~~(8x^2-5)^{-3}$$
$$ y= 4(2x-5)^3~f'~~(8x^2-5)^{-3}-3(2x-5)^4~~(8x^2-5)^{-4}~g'$$
$$ y= 4(2x-5)^3~(2)~~(8x^2-5)^{-3}-3(2x-5)^4~~(8x^2-5)^{-4}~(16x)$$

amistre64 · Answer

to clean it up, that looks to arrange as:$$y= 8\left(\frac{2x-5}{8x^2-5}\right)^3~-~48x~\left(\frac{2x-5}{8x^2-5}\right)^4$$

amistre64 · Answer

you simply have to keep better track of whats going on; dont try to formulate it all at once, but work it out in pieces

anonymous · Answer

so your answer is somehow the same as my book answer