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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\int x^{- \frac{1}{4}} d x

\frac{d y}{d x} = 8 x^{3} y - 8 x y

\int x 2 x d x

\int x (x + 7)^{- \frac{3}{2}} d x

\int_{- 2}^{3} \frac{2}{x ^{4}} d x

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