What is the derivative of y= 4/(1+x^2) ?

The derivative of y= 4/(1+x^2) is -(8x)/((1+x^2)^2)

What is the first derivative of y= 4/(1+x^2) ?

The first derivative of y= 4/(1+x^2) is -(8x)/((1+x^2)^2)

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derivative of

y = \frac{4}{1 + x ^{2}}

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

= 4 \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

= 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}}

\frac{d ^{4} y}{d x ^{4}} + 3 y = 0

x \to 2 lim (- 1)

\int \frac{7 x}{x + 16} d x

\int_{1}^{5} \frac{x ^{2} + 4}{6 x - x ^{2}} d x

f (x) = x^{2} - 1, at x = 4

What is the derivative of y= 4/(1+x^2) ?
The derivative of y= 4/(1+x^2) is -(8x)/((1+x^2)^2)
What is the first derivative of y= 4/(1+x^2) ?
The first derivative of y= 4/(1+x^2) is -(8x)/((1+x^2)^2)