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

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

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

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

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derivative of

f (x) = x^{2} + 9

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

Solution steps

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

Apply the chain rule:

\frac{1}{2 x ^{2} + 9} \frac{d}{d x} (x^{2} + 9)

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

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

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

Simplify

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

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

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

f (x) = 5 x + \frac{3}{x}

(4 z + e^{- z^{2}})^{5}

x \to a lim (2 x)

\int_{- \infty}^{0} x e^{2 x} d x

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