What is the derivative of 1/9 (x^2+6^{3/2}) ?

The derivative of 1/9 (x^2+6^{3/2}) is (x(x^2+6)^{1/2})/3

What is the first derivative of 1/9 (x^2+6^{3/2}) ?

The first derivative of 1/9 (x^2+6^{3/2}) is (x(x^2+6)^{1/2})/3

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\frac{d}{d x} (\frac{1}{9} (x^{2} + 6)^{\frac{3}{2}})

\frac{x ( x ^{2} + 6 ) ^{\frac{1}{2}}}{3}

Solution steps

\frac{d}{d x} (\frac{1}{9} (x^{2} + 6)^{\frac{3}{2}})

Take the constant out:

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

= \frac{1}{9} \frac{d}{d x} ((x^{2} + 6)^{\frac{3}{2}})

Apply the chain rule:

\frac{3 ( x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \frac{d}{d x} (x^{2} + 6)

= \frac{3 ( x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \frac{d}{d x} (x^{2} + 6)

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

= \frac{1}{9} \cdot \frac{3 ( x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \cdot 2 x

Simplify

\frac{1}{9} \cdot \frac{3 ( x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \cdot 2 x : \frac{x ( x ^{2} + 6 ) ^{\frac{1}{2}}}{3}

= \frac{x ( x ^{2} + 6 ) ^{\frac{1}{2}}}{3}

y^{^{'}} + \frac{1}{x} y = 9 x^{2}

\frac{\partial}{\partial x} (x^{2} ln (y + 1))

3 x, x

\int \frac{sec ^{2} ( 3 x )}{1 + tan ( 3 x )} d x

y = x^{2} - 3 x - 5, (2, - 7)

What is the derivative of 1/9 (x^2+6^{3/2}) ?
The derivative of 1/9 (x^2+6^{3/2}) is (x(x^2+6)^{1/2})/3
What is the first derivative of 1/9 (x^2+6^{3/2}) ?
The first derivative of 1/9 (x^2+6^{3/2}) is (x(x^2+6)^{1/2})/3