How do you differentiate $f (x) = {(4 x^{3} + 2 x)}^{- 4}$ $f(x) = (4x^3 + 2x) ^ - 4$ ?

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How do you differentiate $f (x) = {(4 x^{3} + 2 x)}^{- 4}$ $f(x) = (4x^3 + 2x) ^ - 4$ ?

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Answer 1 · 2016-06-09T12:59:27

$\frac{- 8 (6 x^{2} + 1)}{{(4 x^{3} + 2 x)}^{5}}$ $(-8(6x^2+1))/(4x^3+2x)^5$

Explanation:

differentiate f(x) using the $chain rule$ $color(blue)"chain rule"$

f(x) = g(h(x)) then f'(x) = g'(h(x)).h'(x)
$-------------------------------------------$ $"-------------------------------------------"$

$g(h(x))=(4x^3+2x)^-4rArrg'(h(x))=-4(4x^3+2x)^-5$

$h(x)=4x^3+2xrArrh'(x)=12x^2+2=2(6x^2+1)$
$"-------------------------------------------------------------------------"$
Substitute these values into f'(x)

$f'(x)=-4(4x^3+2x)^-5 .2(6x^2+1)$

$=(-8(6x^2+1))/(4x^3+2x)^5$

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How do you differentiate $f (x) = {(4 x^{3} + 2 x)}^{- 4}$ $f(x) = (4x^3 + 2x) ^ - 4$ ?

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Explanation:

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How do you differentiate f(x) = (4x^3 + 2x) ^ - 4f(x)=(4x3+2x)−4f(x) = (4x^3 + 2x) ^ - 4?

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How do you differentiate $f (x) = {(4 x^{3} + 2 x)}^{- 4}$ $f(x) = (4x^3 + 2x) ^ - 4$ ?