What is the derivative of f(x)=(14x+4x^2)/((7+4x)^2) ?

The derivative of f(x)=(14x+4x^2)/((7+4x)^2) is (98)/((7+4x)^3)

What is the first derivative of f(x)=(14x+4x^2)/((7+4x)^2) ?

The first derivative of f(x)=(14x+4x^2)/((7+4x)^2) is (98)/((7+4x)^3)

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derivative of

f (x) = \frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}}

\frac{98}{( 7 + 4 x ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 14 x + 4 x ^{2} ) ( 7 + 4 x ) ^{2} - \frac{d}{d x} ( ( 7 + 4 x ) ^{2} ) ( 14 x + 4 x ^{2} )}{( ( 7 + 4 x ) ^{2} ) ^{2}}

\frac{d}{d x} (14 x + 4 x^{2}) = 14 + 8 x

\frac{d}{d x} ((7 + 4 x)^{2}) = 8 (7 + 4 x)

= \frac{( 14 + 8 x ) ( 7 + 4 x ) ^{2} - 8 ( 7 + 4 x ) ( 14 x + 4 x ^{2} )}{( ( 7 + 4 x ) ^{2} ) ^{2}}

Simplify

\frac{( 14 + 8 x ) ( 7 + 4 x ) ^{2} - 8 ( 7 + 4 x ) ( 14 x + 4 x ^{2} )}{( ( 7 + 4 x ) ^{2} ) ^{2}} : \frac{98}{( 7 + 4 x ) ^{3}}

= \frac{98}{( 7 + 4 x ) ^{3}}

\frac{\partial}{\partial x} (- 2 k l (ln (\frac{y}{- x - y})))

\int \frac{1}{6 - x} d x

g (x) = 2 x - 3

\int 14 θ sec^{2} (θ) d θ

\frac{d}{d x} (9 - x^{2} - y^{2})

What is the derivative of f(x)=(14x+4x^2)/((7+4x)^2) ?
The derivative of f(x)=(14x+4x^2)/((7+4x)^2) is (98)/((7+4x)^3)
What is the first derivative of f(x)=(14x+4x^2)/((7+4x)^2) ?
The first derivative of f(x)=(14x+4x^2)/((7+4x)^2) is (98)/((7+4x)^3)