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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\frac{d}{d x} (\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_{0}^{1} 2 + 1 - x^{2} d x

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

\int sin (t^{2}) d t

f (x) = 4 x^{2} - 5 x, at x = 3

f (x) = - 4 x

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)