What is the derivative of 1/27 (9x^2+6^{3/2}) ?

The derivative of 1/27 (9x^2+6^{3/2}) is x(9x^2+6)^{1/2}

What is the first derivative of 1/27 (9x^2+6^{3/2}) ?

The first derivative of 1/27 (9x^2+6^{3/2}) is x(9x^2+6)^{1/2}

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\frac{d}{d x} (\frac{1}{27} (9 x^{2} + 6)^{\frac{3}{2}})

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

Solution steps

\frac{d}{d x} (\frac{1}{27} (9 x^{2} + 6)^{\frac{3}{2}})

Take the constant out:

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

= \frac{1}{27} \frac{d}{d x} ((9 x^{2} + 6)^{\frac{3}{2}})

Apply the chain rule:

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

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

\frac{d}{d x} (9 x^{2} + 6) = 18 x

= \frac{1}{27} \cdot \frac{3 ( 9 x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \cdot 18 x

Simplify

\frac{1}{27} \cdot \frac{3 ( 9 x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \cdot 18 x : x (9 x^{2} + 6)^{\frac{1}{2}}

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

y^{^{'}} + 9 y = t + e^{- 5 t}

(x - y) d x + x d y = 0

\frac{d}{d x} (- 4 sec (x) + 2 csc (x))

\frac{4 x}{x ^{2} + 1}, \frac{2 x}{25}

\int_{0}^{\infty} x^{2} e^{- \frac{x}{2}} d x

What is the derivative of 1/27 (9x^2+6^{3/2}) ?
The derivative of 1/27 (9x^2+6^{3/2}) is x(9x^2+6)^{1/2}
What is the first derivative of 1/27 (9x^2+6^{3/2}) ?
The first derivative of 1/27 (9x^2+6^{3/2}) is x(9x^2+6)^{1/2}