What is the derivative of (10)/(x^2+1) ?

The derivative of (10)/(x^2+1) is -(20x)/((x^2+1)^2)

What is the first derivative of (10)/(x^2+1) ?

The first derivative of (10)/(x^2+1) is -(20x)/((x^2+1)^2)

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derivative of

\frac{10}{x ^{2} + 1}

- \frac{20 x}{( x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{10}{x ^{2} + 1})

Take the constant out:

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

= 10 \frac{d}{d x} (\frac{1}{x ^{2} + 1})

Apply exponent rule:

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

= 10 \frac{d}{d x} ((x^{2} + 1)^{- 1})

Apply the chain rule:

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

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

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

= 10 (- \frac{1}{( x ^{2} + 1 ) ^{2}} \cdot 2 x)

Simplify

10 (- \frac{1}{( x ^{2} + 1 ) ^{2}} \cdot 2 x) : - \frac{20 x}{( x ^{2} + 1 ) ^{2}}

= - \frac{20 x}{( x ^{2} + 1 ) ^{2}}

x \to 1 lim (\frac{x - 3}{x - 1})

y^{^{'}} + \frac{t}{t ^{2} - 16} y = \frac{e ^{t}}{t - 9}, y (- 7) = - 0.5

\frac{d y}{d x} = x (1 + x) (1 + y)

\int_{0}^{2} ln (x^{2}) d x

\frac{\partial}{\partial x} (\frac{1}{2} (a + b) h)

What is the derivative of (10)/(x^2+1) ?
The derivative of (10)/(x^2+1) is -(20x)/((x^2+1)^2)
What is the first derivative of (10)/(x^2+1) ?
The first derivative of (10)/(x^2+1) is -(20x)/((x^2+1)^2)