What is the derivative of (16/(x(4-x^2))) ?

The derivative of (16/(x(4-x^2))) is -(16(4-3x^2))/(x^2(4-x^2)^2)

What is the first derivative of (16/(x(4-x^2))) ?

The first derivative of (16/(x(4-x^2))) is -(16(4-3x^2))/(x^2(4-x^2)^2)

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\frac{d}{d x} (\frac{16}{x ( 4 - x ^{2} )})

- \frac{16 ( 4 - 3 x ^{2} )}{x ^{2} ( 4 - x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{16}{x ( 4 - x ^{2} )})

Take the constant out:

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

= 16 \frac{d}{d x} (\frac{1}{x ( 4 - x ^{2} )})

Apply exponent rule:

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

= 16 \frac{d}{d x} ((x (4 - x^{2}))^{- 1})

Apply the chain rule:

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

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

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

= 16 (- \frac{1}{( x ( 4 - x ^{2} ) ) ^{2}} (4 - 3 x^{2}))

Simplify

16 (- \frac{1}{( x ( 4 - x ^{2} ) ) ^{2}} (4 - 3 x^{2})) : - \frac{16 ( 4 - 3 x ^{2} )}{x ^{2} ( 4 - x ^{2} ) ^{2}}

= - \frac{16 ( 4 - 3 x ^{2} )}{x ^{2} ( 4 - x ^{2} ) ^{2}}

\int_{2}^{t} \frac{1}{5 x ^{5}} d x

\int \frac{1}{7 + 6 x - x ^{2}} d x

n = 1 \sum \infty (sin (1))^{n}

\int - 3 x d x

\frac{s - 2}{( s + 1 ) ( s - 3 )}

What is the derivative of (16/(x(4-x^2))) ?
The derivative of (16/(x(4-x^2))) is -(16(4-3x^2))/(x^2(4-x^2)^2)
What is the first derivative of (16/(x(4-x^2))) ?
The first derivative of (16/(x(4-x^2))) is -(16(4-3x^2))/(x^2(4-x^2)^2)