What is the derivative of f(x)=(x^2)/((3+x)) ?

The derivative of f(x)=(x^2)/((3+x)) is (6x+x^2)/((3+x)^2)

What is the first derivative of f(x)=(x^2)/((3+x)) ?

The first derivative of f(x)=(x^2)/((3+x)) is (6x+x^2)/((3+x)^2)

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derivative of

f (x) = \frac{x ^{2}}{( 3 + x )}

\frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (3 + x) = 1

= \frac{2 x ( 3 + x ) - 1 \cdot x ^{2}}{( 3 + x ) ^{2}}

Simplify

\frac{2 x ( 3 + x ) - 1 \cdot x ^{2}}{( 3 + x ) ^{2}} : \frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

= \frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

e^{\frac{1}{6} x^{6}}, 0

x \to - 6 - lim (- \frac{1}{x + 6})

\int_{1}^{3} (3 - x)^{2} d x

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

\int \frac{x + 1}{e ^{x}} d x

What is the derivative of f(x)=(x^2)/((3+x)) ?
The derivative of f(x)=(x^2)/((3+x)) is (6x+x^2)/((3+x)^2)
What is the first derivative of f(x)=(x^2)/((3+x)) ?
The first derivative of f(x)=(x^2)/((3+x)) is (6x+x^2)/((3+x)^2)