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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d y} (2 ln (3 x + 4 y))

x \to 0 lim (\frac{4 cos ( x - 4 )}{x})

\int_{0}^{1} \frac{x + 1}{x + 3} d x

x \to - 8 lim (\frac{x ^{2} + 7}{x + 8})

(- 4, - 1), (3, 6)

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