What is the derivative of f(x)=(8x)/(x^2+1) ?

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

What is the first derivative of f(x)=(8x)/(x^2+1) ?

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

= 8 \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}}

= 8 \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

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

Simplify

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

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

y = x^{2} ln (x^{2})

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

ln (x + 3)

\int \frac{e ^{\frac{2}{x}}}{x ^{2}} d x

y^{^{'}} + 6 y = 4 x^{3} e^{- 6 x}

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