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

The derivative of f(x)=(2x)/(sqrt(x^2+9)) is (18)/((x^2+9)sqrt(x^2+9))

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

The first derivative of f(x)=(2x)/(sqrt(x^2+9)) is (18)/((x^2+9)sqrt(x^2+9))

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derivative of

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

\frac{18}{( x ^{2} + 9 ) x ^{2} + 9}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \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}}

= 2 \cdot \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}

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

Simplify

2 \cdot \frac{1 \cdot x ^{2} + 9 - \frac{x}{x ^{2} + 9} x}{( x ^{2} + 9 ) ^{2}} : \frac{18}{( x ^{2} + 9 ) x ^{2} + 9}

= \frac{18}{( x ^{2} + 9 ) x ^{2} + 9}

\int I n^{2} x d x

f (x) = 4 e^{5 x^{2}} (5 - x)^{7}

\frac{\partial}{\partial x} ((x + y) e^{y^{2}})

\int (sin (x^{2})) d x

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

What is the derivative of f(x)=(2x)/(sqrt(x^2+9)) ?
The derivative of f(x)=(2x)/(sqrt(x^2+9)) is (18)/((x^2+9)sqrt(x^2+9))
What is the first derivative of f(x)=(2x)/(sqrt(x^2+9)) ?
The first derivative of f(x)=(2x)/(sqrt(x^2+9)) is (18)/((x^2+9)sqrt(x^2+9))