What is the derivative of y= x/(sqrt(x^2+9)) ?

The derivative of y= x/(sqrt(x^2+9)) is 9/((x^2+9)sqrt(x^2+9))

What is the first derivative of y= x/(sqrt(x^2+9)) ?

The first derivative of y= x/(sqrt(x^2+9)) is 9/((x^2+9)sqrt(x^2+9))

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derivative of

y = \frac{x}{x ^{2} + 9}

\frac{9}{( x ^{2} + 9 ) x ^{2} + 9}

Solution steps

\frac{d}{d x} (\frac{x}{x ^{2} + 9})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d x}{d x} x ^{2} + 9 - \frac{d}{d x} ( x ^{2} + 9 ) x}{( x ^{2} + 9 ) ^{2}}

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

\frac{d}{d x} (x^{2} + 9) = \frac{x}{x ^{2} + 9}

= \frac{1 \cdot x ^{2} + 9 - \frac{x}{x ^{2} + 9} x}{( x ^{2} + 9 ) ^{2}}

Simplify

\frac{1 \cdot x ^{2} + 9 - \frac{x}{x ^{2} + 9} x}{( x ^{2} + 9 ) ^{2}} : \frac{9}{( x ^{2} + 9 ) x ^{2} + 9}

= \frac{9}{( x ^{2} + 9 ) x ^{2} + 9}

x \to \infty lim (1 + \frac{2}{x})

\int (5 x - 2) d x

\int 4 x - 7 d x

y = 0, y = x^{2}, y^{2} = x

x^{2} arctan (9 x)

What is the derivative of y= x/(sqrt(x^2+9)) ?
The derivative of y= x/(sqrt(x^2+9)) is 9/((x^2+9)sqrt(x^2+9))
What is the first derivative of y= x/(sqrt(x^2+9)) ?
The first derivative of y= x/(sqrt(x^2+9)) is 9/((x^2+9)sqrt(x^2+9))