How do you find the first, second derivative for #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

#color(blue)(d/dx(x^n) = n * x^(n-1))#

The first derivative will be equal to

#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)#