How do you find the first, second derivative for $3 x^{\frac{2}{3}} - x^{2}$ $3x^(2/3)-x^2$ ?

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Answer 1 · 2015-08-04T22:13:42

You can use the power rule.

This function's first and second derivates can be found by using the power rule

$\frac{d}{d x} (x^{n}) = n \cdot x^{n - 1}$ $color(blue)(d/dx(x^n) = n * x^(n-1))$

The first derivative will be equal to

$\frac{d}{d x} (f (x)) = [\frac{d}{d x} (3 x^{\frac{2}{3}})] - \frac{d}{d x} (x^{2})$ $d/dx(f(x)) = [d/dx(3x^(2/3))] - d/dx(x^2)$

$f^' = color(red)(cancel(color(black)(3))) * 2/color(red)(cancel(color(black)(3))) * x^(2/3-1) - 2x^(2-1)$

$f^' = color(green)(2 * x^(-1/3) - 2x)$

The second derivative will be equal to

$d/dx(f^'(x)) = [d/dx(2x^(-1/3))] - d/dx(2x)$

$f^('') = 2 * (-1/3) * x^(-1/3-1) - 2$

$f^('') = color(green)(-2/3 * x^(-4/3) - 2)$

How do you find the first, second derivative for 3x^(2/3)-x^23x23−x23x^(2/3)-x^2?