How do you find the derivative of $y = \sqrt[3]{5 x^{3} - 3 x^{2} + 6}$ $y=root3(5x^3 -3x^2 +6)$ ?

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Answer 1 · 2015-10-15T11:31:47

$\frac{d y}{d x} = \frac{15 x^{2} - 6 x}{3 \sqrt[3]{{(5 x^{3} - 3 x^{2} + 6)}^{2}}}$ $dy/dx=(15x^2-6x)/(3root(3)((5x^3-3x^2+6)^2)$

Using laws of indices and surds we may write this function as

$y = {(5 x^{3} - 3 x^{2} + 6)}^{\frac{1}{3}}$ $y=(5x^3-3x^2+6)^(1/3)$

Now we may use the power rule to differentiate it as follows :

$\frac{d y}{d x} = \frac{1}{3} {(5 x^{3} - 3 x^{2} + 6)}^{- \frac{2}{3}} \cdot (15 x^{2} - 6 x)$ $dy/dx=1/3(5x^3-3x^2+6)^(-2/3)*(15x^2-6x)$

$= \frac{15 x^{2} - 6 x}{3 \sqrt[3]{{(5 x^{3} - 3 x^{2} + 6)}^{2}}}$ $=(15x^2-6x)/(3root(3)((5x^3-3x^2+6)^2)$

How do you find the derivative of y=3√5x3−3x2+6y=root3(5x^3 -3x^2 +6) ?