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

The derivative of 1/2 (6t+3)^{-1/2} is -3/(2(6t+3)^{3/2)}

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x^{2} - 7 x - 26, - x^{2} + 3 x + 2

y^{^{'}} = x (x - 1) (x - 5)

\int \frac{y}{1 + x ^{2} y ^{2}} d x

\int \frac{- cos ( n x )}{n} d x

t \frac{d y}{d t} = t^{3} + 24 t^{3} y

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