What is the derivative of (18x+4x^2)/((9+4x)^2) ?

The derivative of (18x+4x^2)/((9+4x)^2) is (162)/((9+4x)^3)

What is the first derivative of (18x+4x^2)/((9+4x)^2) ?

The first derivative of (18x+4x^2)/((9+4x)^2) is (162)/((9+4x)^3)

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derivative of

\frac{18 x + 4 x ^{2}}{( 9 + 4 x ) ^{2}}

\frac{162}{( 9 + 4 x ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{18 x + 4 x ^{2}}{( 9 + 4 x ) ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 18 x + 4 x ^{2} ) ( 9 + 4 x ) ^{2} - \frac{d}{d x} ( ( 9 + 4 x ) ^{2} ) ( 18 x + 4 x ^{2} )}{( ( 9 + 4 x ) ^{2} ) ^{2}}

\frac{d}{d x} (18 x + 4 x^{2}) = 18 + 8 x

\frac{d}{d x} ((9 + 4 x)^{2}) = 8 (9 + 4 x)

= \frac{( 18 + 8 x ) ( 9 + 4 x ) ^{2} - 8 ( 9 + 4 x ) ( 18 x + 4 x ^{2} )}{( ( 9 + 4 x ) ^{2} ) ^{2}}

Simplify

\frac{( 18 + 8 x ) ( 9 + 4 x ) ^{2} - 8 ( 9 + 4 x ) ( 18 x + 4 x ^{2} )}{( ( 9 + 4 x ) ^{2} ) ^{2}} : \frac{162}{( 9 + 4 x ) ^{3}}

= \frac{162}{( 9 + 4 x ) ^{3}}

\int e^{- x} cos (a x) d x

\int 2 x e^{- y} d x

\frac{d}{d x} (4 cos (x) + 2 sin (x))

x \to \infty lim (π^{x})

f (x) = \frac{2}{x + 1}, (4, \frac{2}{5})

What is the derivative of (18x+4x^2)/((9+4x)^2) ?
The derivative of (18x+4x^2)/((9+4x)^2) is (162)/((9+4x)^3)
What is the first derivative of (18x+4x^2)/((9+4x)^2) ?
The first derivative of (18x+4x^2)/((9+4x)^2) is (162)/((9+4x)^3)