What is the derivative of 4/((9+x^2)) ?

The derivative of 4/((9+x^2)) is -(8x)/((9+x^2)^2)

What is the first derivative of 4/((9+x^2)) ?

The first derivative of 4/((9+x^2)) is -(8x)/((9+x^2)^2)

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derivative of

\frac{4}{( 9 + x ^{2} )}

- \frac{8 x}{( 9 + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4}{9 + x ^{2}})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{1}{9 + x ^{2}})

Apply exponent rule:

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

= 4 \frac{d}{d x} ((9 + x^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( 9 + x ^{2} ) ^{2}} \frac{d}{d x} (9 + x^{2})

= - \frac{1}{( 9 + x ^{2} ) ^{2}} \frac{d}{d x} (9 + x^{2})

\frac{d}{d x} (9 + x^{2}) = 2 x

= 4 (- \frac{1}{( 9 + x ^{2} ) ^{2}} \cdot 2 x)

Simplify

4 (- \frac{1}{( 9 + x ^{2} ) ^{2}} \cdot 2 x) : - \frac{8 x}{( 9 + x ^{2} ) ^{2}}

= - \frac{8 x}{( 9 + x ^{2} ) ^{2}}

d er i v a t i v e (x + 1)^{3} (5 - x)

d er i v a t i v e y = (x^{2} + 1) \cdot (x + 1)

d er i v a t i v e \frac{( 2 + x )}{( 2 - x )}

d er i v a t i v e arctan (x^{2} + 1)

d er i v a t i v e y = \frac{( x - 3 ) ^{2} - 4}{x ^{2} + 2}

What is the derivative of 4/((9+x^2)) ?
The derivative of 4/((9+x^2)) is -(8x)/((9+x^2)^2)
What is the first derivative of 4/((9+x^2)) ?
The first derivative of 4/((9+x^2)) is -(8x)/((9+x^2)^2)