What is the derivative of g(x)=-9/((6x+8)^{3/2)} ?

The derivative of g(x)=-9/((6x+8)^{3/2)} is (81)/((6x+8)^{5/2)}

What is the first derivative of g(x)=-9/((6x+8)^{3/2)} ?

The first derivative of g(x)=-9/((6x+8)^{3/2)} is (81)/((6x+8)^{5/2)}

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derivative of

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

\frac{81}{( 6 x + 8 ) ^{\frac{5}{2}}}

Solution steps

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

Take the constant out:

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

= - 9 \frac{d}{d x} (\frac{1}{( 6 x + 8 ) ^{\frac{3}{2}}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

- 9 (- \frac{3}{2 ( 6 x + 8 ) ^{\frac{5}{2}}} \cdot 6) : \frac{81}{( 6 x + 8 ) ^{\frac{5}{2}}}

= \frac{81}{( 6 x + 8 ) ^{\frac{5}{2}}}

\int 5^{3 x} d x

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What is the derivative of g(x)=-9/((6x+8)^{3/2)} ?
The derivative of g(x)=-9/((6x+8)^{3/2)} is (81)/((6x+8)^{5/2)}
What is the first derivative of g(x)=-9/((6x+8)^{3/2)} ?
The first derivative of g(x)=-9/((6x+8)^{3/2)} is (81)/((6x+8)^{5/2)}