What is the derivative of f(x)= 5/(1+x^2) ?

The derivative of f(x)= 5/(1+x^2) is -(10x)/((1+x^2)^2)

What is the first derivative of f(x)= 5/(1+x^2) ?

The first derivative of f(x)= 5/(1+x^2) is -(10x)/((1+x^2)^2)

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derivative of

f (x) = \frac{5}{1 + x ^{2}}

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

Solution steps

\frac{d}{d x} (\frac{5}{1 + x ^{2}})

Take the constant out:

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

= 5 \frac{d}{d x} (\frac{1}{1 + x ^{2}})

Apply exponent rule:

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

= 5 \frac{d}{d x} ((1 + x^{2})^{- 1})

Apply the chain rule:

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

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

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

= 5 (- \frac{1}{( 1 + x ^{2} ) ^{2}} \cdot 2 x)

Simplify

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

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

n = 1 \sum \infty n e^{- 4 n}

y = x^{4} + 9 x^{2} - x

\int \frac{1}{2} sec^{2} (5 x) d x

x + \frac{2}{x}

What is the derivative of f(x)= 5/(1+x^2) ?
The derivative of f(x)= 5/(1+x^2) is -(10x)/((1+x^2)^2)
What is the first derivative of f(x)= 5/(1+x^2) ?
The first derivative of f(x)= 5/(1+x^2) is -(10x)/((1+x^2)^2)