What is the derivative of 1/2 (8t+3)^{-1/2} ?

The derivative of 1/2 (8t+3)^{-1/2} is -2/((8t+3)^{3/2)}

What is the first derivative of 1/2 (8t+3)^{-1/2} ?

The first derivative of 1/2 (8t+3)^{-1/2} is -2/((8t+3)^{3/2)}

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (f (x) \cdot f (x) \cdot f (x))

y^{^{'}} - \frac{8}{x} y = y^{3} \frac{1}{x ^{12}}

y = x^{2}, x = y^{2}

3 x y^{^{'}} - 15 y = x^{- 1}

n = 9 \sum \infty (\frac{1}{7})^{n}

What is the derivative of 1/2 (8t+3)^{-1/2} ?
The derivative of 1/2 (8t+3)^{-1/2} is -2/((8t+3)^{3/2)}
What is the first derivative of 1/2 (8t+3)^{-1/2} ?
The first derivative of 1/2 (8t+3)^{-1/2} is -2/((8t+3)^{3/2)}