Expert answers to "Solve the derivatives. v=(z4−2z+1)32"

nicekikah

Answered question

2021-10-03

Solve the derivatives.

v = {(z^{4} - 2 z + 1)}^{\frac{3}{2}}

lamusesamuset

Skilled2021-10-04Added 93 answers

Step 1
Given:

v = {(z^{4} - 2 z + 1)}^{\frac{3}{2}}

Step 2

v = {(z^{4} - 2 z + 1)}^{\frac{3}{2}}

Differentiate with respect to z

\frac{d v}{d z} = \frac{d}{d z} [{(z^{4} - 2 z + 1)}^{\frac{3}{2}}]

Apply chain rule:

\frac{d}{d x} (f (g (x))) = \frac{d}{d u} f (u) \frac{d}{d x} g (x) w i t h u = (z^{4} - 2 z + 1)

\frac{d v}{d z} = \frac{d}{d u} u^{\frac{3}{2}} \frac{d}{d x} (z^{4} - 2 z + 1)

Apply power rule:

\frac{d}{d x} (x^{n}) = n x^{n - 1}

\frac{d v}{d z} = \frac{3}{2} u^{\frac{3}{2} - 1} (4 z^{4 - 1} - 2 + 0)

\frac{d v}{d z} = \frac{3}{2} u^{\frac{1}{2}} (4 z^{3} - 2)

Put back:

u = (z^{4} - 2 z + 1)

\frac{d v}{d z} = \frac{3}{2} {(z^{4} - 2 z + 1)}^{\frac{1}{2}} (4 z^{3} - 2)

Jeffrey Jordon

Expert2022-02-01Added 2605 answers

Answer is given below (on video)

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