What is the first derivative of the curve described by $y = \frac{1}{2} \sqrt[3]{x} + \frac{8}{x} + 1$ $y = 1/2root(3)(x) + 8/x + 1$ ?

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Answer 1 · 2016-11-21T02:05:12

First of all, we can rewrite the curve as

$y = \frac{1}{2} x^{\frac{1}{3}} + 8 x^{- 1} + 1$ $y = 1/2x^(1/3) + 8x^(-1) + 1$

Differentiate using the power rule, $(x^n)' = nx^(n - 1)$ .

$dy/dx = 1/3(1/2)x^(-2/3) + (-1)8x^(-2) + 0$

$dy/dx = 1/(6x^(2/3)) - 8/x^2$

Hopefully this helps!

What is the first derivative of the curve described by y = 1/2root(3)(x) + 8/x + 1y=123√x+8x+1y = 1/2root(3)(x) + 8/x + 1?