What is the derivative of (7x/(x^2+4)) ?

The derivative of (7x/(x^2+4)) is (7(-x^2+4))/((x^2+4)^2)

What is the first derivative of (7x/(x^2+4)) ?

The first derivative of (7x/(x^2+4)) is (7(-x^2+4))/((x^2+4)^2)

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\frac{d}{d x} (\frac{7 x}{x ^{2} + 4})

\frac{7 ( - x ^{2} + 4 )}{( x ^{2} + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{7 x}{x ^{2} + 4})

Take the constant out:

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

= 7 \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}}

= 7 \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

= 7 \cdot \frac{1 \cdot ( x ^{2} + 4 ) - 2 xx}{( x ^{2} + 4 ) ^{2}}

Simplify

7 \cdot \frac{1 \cdot ( x ^{2} + 4 ) - 2 xx}{( x ^{2} + 4 ) ^{2}} : \frac{7 ( - x ^{2} + 4 )}{( x ^{2} + 4 ) ^{2}}

= \frac{7 ( - x ^{2} + 4 )}{( x ^{2} + 4 ) ^{2}}

\frac{1}{6} x^{3} + \frac{1}{2 x}

\frac{d}{d x} (4 + 3 f (x))

x \to 0 + lim (ln (\frac{9}{x ^{3}}))

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

e^{4 t}

What is the derivative of (7x/(x^2+4)) ?
The derivative of (7x/(x^2+4)) is (7(-x^2+4))/((x^2+4)^2)
What is the first derivative of (7x/(x^2+4)) ?
The first derivative of (7x/(x^2+4)) is (7(-x^2+4))/((x^2+4)^2)