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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

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

Simplify

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

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

\int \frac{28}{1 - cos ( 4 x )} d x

y^{^{'}} = y + x, y (0) = 1

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

\int sin (t) e^{- \frac{2}{3} t} d t

\int_{1}^{2} x^{2} e^{x^{3} - 1} d x

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