What is the derivative of y=24(x^3+6x)^{3/8} ?

The derivative of y=24(x^3+6x)^{3/8} is (9(3x^2+6))/((x^3+6x)^{5/8)}

What is the first derivative of y=24(x^3+6x)^{3/8} ?

The first derivative of y=24(x^3+6x)^{3/8} is (9(3x^2+6))/((x^3+6x)^{5/8)}

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derivative of

y = 24 (x^{3} + 6 x)^{\frac{3}{8}}

\frac{9 ( 3 x ^{2} + 6 )}{( x ^{3} + 6 x ) ^{\frac{5}{8}}}

Solution steps

\frac{d}{d x} (24 (x^{3} + 6 x)^{\frac{3}{8}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 24 \frac{d}{d x} ((x^{3} + 6 x)^{\frac{3}{8}})

Apply the chain rule:

\frac{3}{8 ( x ^{3} + 6 x ) ^{\frac{5}{8}}} \frac{d}{d x} (x^{3} + 6 x)

= \frac{3}{8 ( x ^{3} + 6 x ) ^{\frac{5}{8}}} \frac{d}{d x} (x^{3} + 6 x)

\frac{d}{d x} (x^{3} + 6 x) = 3 x^{2} + 6

= 24 \cdot \frac{3}{8 ( x ^{3} + 6 x ) ^{\frac{5}{8}}} (3 x^{2} + 6)

Simplify

24 \cdot \frac{3}{8 ( x ^{3} + 6 x ) ^{\frac{5}{8}}} (3 x^{2} + 6) : \frac{9 ( 3 x ^{2} + 6 )}{( x ^{3} + 6 x ) ^{\frac{5}{8}}}

= \frac{9 ( 3 x ^{2} + 6 )}{( x ^{3} + 6 x ) ^{\frac{5}{8}}}

\int x (x - 1) (3 - x) d x

\int_{2}^{3} \frac{x}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} + y ^{4}})

\int \frac{e ^{6 x}}{e ^{12 x} + 49} d x

What is the derivative of y=24(x^3+6x)^{3/8} ?
The derivative of y=24(x^3+6x)^{3/8} is (9(3x^2+6))/((x^3+6x)^{5/8)}
What is the first derivative of y=24(x^3+6x)^{3/8} ?
The first derivative of y=24(x^3+6x)^{3/8} is (9(3x^2+6))/((x^3+6x)^{5/8)}