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

The derivative of f(x)=(6x+x^2)/((3+x)^2) is (18)/((3+x)^3)

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

The first derivative of f(x)=(6x+x^2)/((3+x)^2) is (18)/((3+x)^3)

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derivative of

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

\frac{18}{( 3 + x ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 6 + 2 x ) ( 3 + x ) ^{2} - 2 ( 3 + x ) ( 6 x + x ^{2} )}{( ( 3 + x ) ^{2} ) ^{2}} : \frac{18}{( 3 + x ) ^{3}}

= \frac{18}{( 3 + x ) ^{3}}

\int \frac{3 - 18 x}{1 - 9 x ^{2}} d x

\frac{d}{d x} (ln (x - 5))

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

2 (t - \frac{π}{4}) sin (4 t)

3 x^{2} + 3 x

What is the derivative of f(x)=(6x+x^2)/((3+x)^2) ?
The derivative of f(x)=(6x+x^2)/((3+x)^2) is (18)/((3+x)^3)
What is the first derivative of f(x)=(6x+x^2)/((3+x)^2) ?
The first derivative of f(x)=(6x+x^2)/((3+x)^2) is (18)/((3+x)^3)