How do you differentiate $y=(2x-5)^4(8x^2-5)^-3$ ?

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Answer 1 · 2017-01-23T06:07:37

$y'=(8(2x-5)^3(-4x^2+30x-5))/((8x^2-5)^4)$

You would apply the rule:

$y=f(x)g(x)->y'=color(red)(f'(x))g(x)+f(x)color(green)(g'(x))$

and the chain rule, since

$f(x)=(2x-5)^4 and g(x)=(8x^2-5)^-3$

Then:

$y'=color(red)(4(2x-5)^3*2)(8x^2-5)^-3+(2x-5)^4*color(green)((-3)(8x^2-5)^-4*16x)$

$=8(2x-5)^3/(8x^2-5)^3-48x(2x-5)^4/(8x^2-5)^4$

$=(8(8x^2-5)(2x-5)^3-48x(2x-5)^4)/((8x^2-5)^4)$

$=(8(2x-5)^3(8x^2-5-12x^2+30x))/((8x^2-5)^4)$

$=(8(2x-5)^3(-4x^2+30x-5))/((8x^2-5)^4)$

How do you differentiate y=(2x-5)^4(8x^2-5)^-3y=(2x-5)^4(8x^2-5)^-3?