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

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

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

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

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derivative of

y = x^{2} + 4

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

Solution steps

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

Apply the chain rule:

\frac{1}{2 x ^{2} + 4} \frac{d}{d x} (x^{2} + 4)

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

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

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

Simplify

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

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

\frac{d y}{d x} = (x - 3) e^{(- 2 y)}

\frac{d}{d x} (\frac{e ^{3 x} sin ( 2 x )}{5 ^{x}})

\int \frac{cot ( x )}{1 - sin ( x )} d x

- 4 t^{2} y^{^{''}} + 5 t y^{^{'}} - 2 y = 0

\int \frac{sec ^{2} ( \frac{1}{x ^{9}} )}{x ^{10}} d x

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