What is the derivative of (6x/(sqrt(x^2+4))) ?

The derivative of (6x/(sqrt(x^2+4))) is (24)/((x^2+4)sqrt(x^2+4))

What is the first derivative of (6x/(sqrt(x^2+4))) ?

The first derivative of (6x/(sqrt(x^2+4))) is (24)/((x^2+4)sqrt(x^2+4))

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\frac{d}{d x} (\frac{6 x}{x ^{2} + 4})

\frac{24}{( x ^{2} + 4 ) x ^{2} + 4}

Solution steps

\frac{d}{d x} (\frac{6 x}{x ^{2} + 4})

Take the constant out:

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

= 6 \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}}

= 6 \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) = \frac{x}{x ^{2} + 4}

= 6 \cdot \frac{1 \cdot x ^{2} + 4 - \frac{x}{x ^{2} + 4} x}{( x ^{2} + 4 ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot x ^{2} + 4 - \frac{x}{x ^{2} + 4} x}{( x ^{2} + 4 ) ^{2}} : \frac{24}{( x ^{2} + 4 ) x ^{2} + 4}

= \frac{24}{( x ^{2} + 4 ) x ^{2} + 4}

\int 99 - 2 x - x^{2} d x

\frac{d y}{d x} = \frac{1}{9} y cos^{2} (y)

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

4 sin (x), at x = \frac{π}{2}

\frac{- 1}{s + 2}

What is the derivative of (6x/(sqrt(x^2+4))) ?
The derivative of (6x/(sqrt(x^2+4))) is (24)/((x^2+4)sqrt(x^2+4))
What is the first derivative of (6x/(sqrt(x^2+4))) ?
The first derivative of (6x/(sqrt(x^2+4))) is (24)/((x^2+4)sqrt(x^2+4))