What is the derivative of y= x/(sqrt(x^2+1)) ?

The derivative of y= x/(sqrt(x^2+1)) is 1/((x^2+1)sqrt(x^2+1))

What is the first derivative of y= x/(sqrt(x^2+1)) ?

The first derivative of y= x/(sqrt(x^2+1)) is 1/((x^2+1)sqrt(x^2+1))

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derivative of

y = \frac{x}{x ^{2} + 1}

\frac{1}{( x ^{2} + 1 ) x ^{2} + 1}

Solution steps

\frac{d}{d x} (\frac{x}{x ^{2} + 1})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{2} + 1) = \frac{x}{x ^{2} + 1}

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

Simplify

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

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

3 x^{2} - \frac{1}{x}

f (x) = x^{2} + 2 x, at x = 34

x \to \infty lim (2^{x})

\int \frac{x ^{2} + 2 x - 1}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x

f (x) = (4 x - 9)^{2}

What is the derivative of y= x/(sqrt(x^2+1)) ?
The derivative of y= x/(sqrt(x^2+1)) is 1/((x^2+1)sqrt(x^2+1))
What is the first derivative of y= x/(sqrt(x^2+1)) ?
The first derivative of y= x/(sqrt(x^2+1)) is 1/((x^2+1)sqrt(x^2+1))