What is the derivative of f(x)=((x^2))/((4+x)) ?

The derivative of f(x)=((x^2))/((4+x)) is (8x+x^2)/((4+x)^2)

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

The first derivative of f(x)=((x^2))/((4+x)) is (8x+x^2)/((4+x)^2)

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derivative of

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

\frac{8 x + x ^{2}}{( 4 + x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 x ( 4 + x ) - 1 \cdot x ^{2}}{( 4 + x ) ^{2}} : \frac{8 x + x ^{2}}{( 4 + x ) ^{2}}

= \frac{8 x + x ^{2}}{( 4 + x ) ^{2}}

\int_{1}^{4} t^{2} ln (4 t) d t

y^{^{'}} + 6 y = t + e^{- 4 t}

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

\frac{\partial}{\partial x} (2 x (x^{2} + 1)^{- 1})

e^{4 x}

What is the derivative of f(x)=((x^2))/((4+x)) ?
The derivative of f(x)=((x^2))/((4+x)) is (8x+x^2)/((4+x)^2)
What is the first derivative of f(x)=((x^2))/((4+x)) ?
The first derivative of f(x)=((x^2))/((4+x)) is (8x+x^2)/((4+x)^2)