How do you differentiate ${([s^{4}] - 8)}^{.25}$ $([s^4] - 8)^ .25$ ?

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Answer 1 · 2017-02-12T09:32:04

${(s^{4} - 8)}^{- \frac{3}{4}} \cdot s^{3}$ $(s^4-8)^(-3/4)*s^3$

Use the power rule to find the derivative of the outside

${(s^{4} - 8)}^{\frac{1}{4}}$ $(s^4-8)^(1/4)$

$(\frac{1}{4}) {(s^{4} - 8)}^{(\frac{1}{4} - 1)}$ $(1/4)(s^4-8)^((1/4-1)$

But then you have to times this by the derivative of the inside which in this case is the derivative of $(s^{4} - 8)$ $(s^4-8)$

$(\frac{1}{4}) {(s^{4} - 8)}^{\frac{1}{4} - 1}$ $(1/4)(s^4-8)^(1/4-1)$ times $4 s^{3}$ $4s^3$

$\frac{1}{4} {(s^{4} - 8)}^{- \frac{3}{4}} \cdot 4 s^{3}$ $1/4(s^4-8)^(-3/4)*4s^3$

${(s^{4} - 8)}^{- \frac{3}{4}} \cdot s^{3}$ $(s^4-8)^(-3/4)*s^3$

How do you differentiate ([s^4] - 8)^ .25([s4]−8).25([s^4] - 8)^ .25?