What is the derivative of (6x+4x^2/((3+4x)^2)) ?

The derivative of (6x+4x^2/((3+4x)^2)) is (18)/((3+4x)^3)

What is the first derivative of (6x+4x^2/((3+4x)^2)) ?

The first derivative of (6x+4x^2/((3+4x)^2)) is (18)/((3+4x)^3)

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\frac{d}{d x} (\frac{6 x + 4 x ^{2}}{( 3 + 4 x ) ^{2}})

\frac{18}{( 3 + 4 x ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{6 x + 4 x ^{2}}{( 3 + 4 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 + 4 x ^{2} ) ( 3 + 4 x ) ^{2} - \frac{d}{d x} ( ( 3 + 4 x ) ^{2} ) ( 6 x + 4 x ^{2} )}{( ( 3 + 4 x ) ^{2} ) ^{2}}

\frac{d}{d x} (6 x + 4 x^{2}) = 6 + 8 x

\frac{d}{d x} ((3 + 4 x)^{2}) = 8 (3 + 4 x)

= \frac{( 6 + 8 x ) ( 3 + 4 x ) ^{2} - 8 ( 3 + 4 x ) ( 6 x + 4 x ^{2} )}{( ( 3 + 4 x ) ^{2} ) ^{2}}

Simplify

\frac{( 6 + 8 x ) ( 3 + 4 x ) ^{2} - 8 ( 3 + 4 x ) ( 6 x + 4 x ^{2} )}{( ( 3 + 4 x ) ^{2} ) ^{2}} : \frac{18}{( 3 + 4 x ) ^{3}}

= \frac{18}{( 3 + 4 x ) ^{3}}

y^{^{'}} + 5 e^{x + y} = 0

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

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

\int x^{3} x^{2} - 9 d x

\frac{d y}{d x} = 2^{x - y}, y (- 3) = - 5

What is the derivative of (6x+4x^2/((3+4x)^2)) ?
The derivative of (6x+4x^2/((3+4x)^2)) is (18)/((3+4x)^3)
What is the first derivative of (6x+4x^2/((3+4x)^2)) ?
The first derivative of (6x+4x^2/((3+4x)^2)) is (18)/((3+4x)^3)