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

The derivative of y= x/(sqrt(x^2+4)) is 4/((x^2+4)sqrt(x^2+4))

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

The first derivative of y= x/(sqrt(x^2+4)) is 4/((x^2+4)sqrt(x^2+4))

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derivative of

y = \frac{x}{x ^{2} + 4}

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

Solution steps

\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}}

= \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}

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

Simplify

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

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

\frac{\partial}{\partial y} (\frac{x y}{x + y + z})

\frac{\partial}{\partial y} (2 x y - y)

f (x) = \frac{7}{x}

\frac{sin ( x )}{cos ( x )}

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

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