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

The derivative of-1/((2x+8^{3/2)}) is 3/((2x+8)^{5/2)}

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

The first derivative of-1/((2x+8^{3/2)}) is 3/((2x+8)^{5/2)}

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (2 x + 8) = 2

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

Simplify

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

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

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

\frac{d}{d x} (\frac{x}{x + 5})

\frac{d}{d t} (cos (2 t))

y^{^{''}} - 2 y^{^{'}} + y = x^{2} - 1

y^{^{'}} + 4 y = e^{- t} - 4

What is the derivative of-1/((2x+8^{3/2)}) ?
The derivative of-1/((2x+8^{3/2)}) is 3/((2x+8)^{5/2)}
What is the first derivative of-1/((2x+8^{3/2)}) ?
The first derivative of-1/((2x+8^{3/2)}) is 3/((2x+8)^{5/2)}