What is (\partial)/(\partial x)(4/(1+x^2)) ?

The answer to (\partial)/(\partial x)(4/(1+x^2)) is -(8x)/((1+x^2)^2)

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(\partial)/(\partial x)(4/(1+x^2))

\frac{\partial}{\partial x} (\frac{4}{1 + x ^{2}})

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

Solution steps

\frac{\partial}{\partial x} (\frac{4}{1 + x ^{2}})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (1 + x^{2}) = 2 x

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

Simplify

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

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

\int \frac{7}{2} x^{2} d x

d er i v a t i v e 4 cos (2 x)

im pl i c i t \frac{d y}{d x}, y = x^{- 10 c o s (x)}

y^{^{'}} = (y + 5) (x + 2)

t an g e n t f (x) = \frac{7}{x}, at x = 4

What is (\partial)/(\partial x)(4/(1+x^2)) ?
The answer to (\partial)/(\partial x)(4/(1+x^2)) is -(8x)/((1+x^2)^2)