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

The derivative of-3/(8x^{5/2}) is (15)/(16x^{7/2)}

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

The first derivative of-3/(8x^{5/2}) is (15)/(16x^{7/2)}

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

\frac{15}{16 x ^{\frac{7}{2}}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Simplify

- \frac{3}{8} (- \frac{5}{2} x^{- \frac{5}{2} - 1}) : \frac{15}{16 x ^{\frac{7}{2}}}

= \frac{15}{16 x ^{\frac{7}{2}}}

\int sec^{2} (5 x) + e^{\frac{x}{2}} d x

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

y^{^{'}} (t) = (y + 1) (y - 3), y (0) = 1

t^{3} \frac{d y}{d t} + 3 t^{2} y = 7 cos (t), y (π) = 0

x \to \infty lim (- \frac{x}{s})

What is the derivative of-3/(8x^{5/2}) ?
The derivative of-3/(8x^{5/2}) is (15)/(16x^{7/2)}
What is the first derivative of-3/(8x^{5/2}) ?
The first derivative of-3/(8x^{5/2}) is (15)/(16x^{7/2)}