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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

= 16 ((x (4 - x^{2}))^{- 1})^{^{'}}

Apply the chain rule:

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

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

(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 (csc^{2} (θ) + 1) d θ

\int tan^{2} (u) d u

\int_{1}^{18} \frac{3}{x} d x

\int 4 e^{- 4 x} sin (4 x) d x

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

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