What is the derivative of-1/(4(1+x^{3/2)}) ?

The derivative of-1/(4(1+x^{3/2)}) is 3/(8(1+x)^{5/2)}

What is the first derivative of-1/(4(1+x^{3/2)}) ?

The first derivative of-1/(4(1+x^{3/2)}) is 3/(8(1+x)^{5/2)}

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\frac{d}{d x} (- \frac{1}{4 ( 1 + x ) ^{\frac{3}{2}}})

\frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}}

Solution steps

\frac{d}{d x} (- \frac{1}{4 ( 1 + x ) ^{\frac{3}{2}}})

Take the constant out:

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

= - \frac{1}{4} \frac{d}{d x} (\frac{1}{( 1 + x ) ^{\frac{3}{2}}})

Apply exponent rule:

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

= - \frac{1}{4} \frac{d}{d x} ((1 + x)^{- \frac{3}{2}})

Apply the chain rule:

- \frac{3}{2 ( 1 + x ) ^{\frac{5}{2}}} \frac{d}{d x} (1 + x)

= - \frac{3}{2 ( 1 + x ) ^{\frac{5}{2}}} \frac{d}{d x} (1 + x)

\frac{d}{d x} (1 + x) = 1

= - \frac{1}{4} (- \frac{3}{2 ( 1 + x ) ^{\frac{5}{2}}} \cdot 1)

Simplify

- \frac{1}{4} (- \frac{3}{2 ( 1 + x ) ^{\frac{5}{2}}} \cdot 1) : \frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}}

= \frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}}

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

n = 1 \sum \infty (x - 4)^{n}

\frac{d}{d x} (ln (x x^{2} - 9))

x \to \infty lim (15)

f (x) = \frac{x}{x + 9}

What is the derivative of-1/(4(1+x^{3/2)}) ?
The derivative of-1/(4(1+x^{3/2)}) is 3/(8(1+x)^{5/2)}
What is the first derivative of-1/(4(1+x^{3/2)}) ?
The first derivative of-1/(4(1+x^{3/2)}) is 3/(8(1+x)^{5/2)}