What is the derivative of 3/(sqrt(6t+8)) ?

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

What is the first derivative of 3/(sqrt(6t+8)) ?

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

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derivative of

\frac{3}{6 t + 8}

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

Solution steps

\frac{d}{d t} (\frac{3}{6 t + 8})

Take the constant out:

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

= 3 \frac{d}{d t} (\frac{1}{6 t + 8})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

cos (x), at x = \frac{π}{4}

\frac{d}{d x} (\frac{g ( x )}{x})

\int - 4 y^{2} d y

\frac{\partial}{\partial x} ((x + y)^{3})

\frac{\partial}{\partial x} (x^{2} yz - x y z^{3})

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