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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 1 - lim (\frac{4}{x ^{3} - 1})

\frac{d}{d x} (3 (x)^{3} + cos (x))

x \to 0 lim (x^{2} csc^{2} (x))

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

\frac{d}{d x} (1 + x^{256} - 1)

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