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

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

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

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

Popular >

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (sin (2 y x))

\frac{d y}{d x} = \frac{2 y}{5 y - 2 x}

\int 3 tan (3 x) d x

\int \frac{2 x ^{2} - 2 x - 1}{x ^{3} - x ^{2}} d x

x \to - 2 lim (\frac{3 x ^{2} + a x + a + 9}{x ^{2} + x - 2})