What is the (-(2x)/((x^2+1)^2))^' ?

The (-(2x)/((x^2+1)^2))^' is -(2(-3x^2+1))/((x^2+1)^3)

What is the first (-(2x)/((x^2+1)^2))^' ?

The first (-(2x)/((x^2+1)^2))^' is -(2(-3x^2+1))/((x^2+1)^3)

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

- \frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

x^{'} = 1

((x^{2} + 1)^{2})^{'} = 4 x (x^{2} + 1)

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

Simplify

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

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

f (x) = \frac{64 x}{x ^{2} + 64}, at x = - 4

x \to 0 + lim (sin (x) n x)

f (x) = x^{2} - 4 x + 3, (1, 0)

\int_{0}^{π} sin (x^{2}) d x

\int cot^{3} (x) d x

What is the (-(2x)/((x^2+1)^2))^' ?
The (-(2x)/((x^2+1)^2))^' is -(2(-3x^2+1))/((x^2+1)^3)
What is the first (-(2x)/((x^2+1)^2))^' ?
The first (-(2x)/((x^2+1)^2))^' is -(2(-3x^2+1))/((x^2+1)^3)