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

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

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

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

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derivative of

\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}}

y^{^{'}} = \frac{6 x ^{3} + y ^{3}}{x y ^{2}}

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

\frac{d y}{d x} = \frac{( 2 x ( y - 1 ) )}{( x ^{2} + 3 )}

y = 2 x^{3} - x^{2} + 2 (1.3)

(10.4) (4.15)

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