What is the derivative of \sqrt[5]{8x^3+9x} ?

The derivative of \sqrt[5]{8x^3+9x} is (24x^2+9)/(5(8x^3+9x)^{4/5)}

What is the first derivative of \sqrt[5]{8x^3+9x} ?

The first derivative of \sqrt[5]{8x^3+9x} is (24x^2+9)/(5(8x^3+9x)^{4/5)}

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\frac{24 x ^{2} + 9}{5 ( 8 x ^{3} + 9 x ) ^{\frac{4}{5}}}

Solution steps

Apply the chain rule:

\frac{1}{5 ( 8 x ^{3} + 9 x ) ^{\frac{4}{5}}} \frac{d}{d x} (8 x^{3} + 9 x)

= \frac{1}{5 ( 8 x ^{3} + 9 x ) ^{\frac{4}{5}}} \frac{d}{d x} (8 x^{3} + 9 x)

\frac{d}{d x} (8 x^{3} + 9 x) = 24 x^{2} + 9

= \frac{1}{5 ( 8 x ^{3} + 9 x ) ^{\frac{4}{5}}} (24 x^{2} + 9)

Simplify

\frac{1}{5 ( 8 x ^{3} + 9 x ) ^{\frac{4}{5}}} (24 x^{2} + 9) : \frac{24 x ^{2} + 9}{5 ( 8 x ^{3} + 9 x ) ^{\frac{4}{5}}}

= \frac{24 x ^{2} + 9}{5 ( 8 x ^{3} + 9 x ) ^{\frac{4}{5}}}

y = 4 x^{3} - 9 x^{2} - 12 x + 12, (1, - 5)

\frac{d}{d x} (\frac{tan ( x ) - 1}{sec ( x )})

\frac{d}{d x} (2 arccos (\frac{x}{4}))

f (x) = 2 sin (x), at x = \frac{π}{3}

What is the derivative of \sqrt[5]{8x^3+9x} ?
The derivative of \sqrt[5]{8x^3+9x} is (24x^2+9)/(5(8x^3+9x)^{4/5)}
What is the first derivative of \sqrt[5]{8x^3+9x} ?
The first derivative of \sqrt[5]{8x^3+9x} is (24x^2+9)/(5(8x^3+9x)^{4/5)}