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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derivative of

\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}

\frac{d y}{d x} = x^{2} y^{2}

f (x) = (ln (x))^{5}, at x = 4

x \to 0 lim (1 + cos (x))

\int 3 x e^{- x^{2}} d x

\int sin (x) ln (cos (x)) d x

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))