What is the derivative of (x(c^2-x^2)/(2.4(c^2+cx))) ?

The derivative of (x(c^2-x^2)/(2.4(c^2+cx))) is (-2cx^3-3c^2x^2+c^4)/(2.4(c^2+cx)^2)

What is the first derivative of (x(c^2-x^2)/(2.4(c^2+cx))) ?

The first derivative of (x(c^2-x^2)/(2.4(c^2+cx))) is (-2cx^3-3c^2x^2+c^4)/(2.4(c^2+cx)^2)

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\frac{d}{d x} (\frac{x ( c ^{2} - x ^{2} )}{2.4 ( c ^{2} + c x )})

\frac{- 2 c x ^{3} - 3 c ^{2} x ^{2} + c ^{4}}{2.4 ( c ^{2} + c x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ( c ^{2} - x ^{2} )}{2.4 ( c ^{2} + c x )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2.4} \frac{d}{d x} (\frac{x ( c ^{2} - x ^{2} )}{c ^{2} + c x})

Apply the Quotient Rule:

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

= \frac{1}{2.4} \cdot \frac{\frac{d}{d x} ( x ( c ^{2} - x ^{2} ) ) ( c ^{2} + c x ) - \frac{d}{d x} ( c ^{2} + c x ) x ( c ^{2} - x ^{2} )}{( c ^{2} + c x ) ^{2}}

\frac{d}{d x} (x (c^{2} - x^{2})) = c^{2} - 3 x^{2}

\frac{d}{d x} (c^{2} + c x) = c

= \frac{1}{2.4} \cdot \frac{( c ^{2} - 3 x ^{2} ) ( c ^{2} + c x ) - c x ( c ^{2} - x ^{2} )}{( c ^{2} + c x ) ^{2}}

Simplify

\frac{1}{2.4} \cdot \frac{( c ^{2} - 3 x ^{2} ) ( c ^{2} + c x ) - c x ( c ^{2} - x ^{2} )}{( c ^{2} + c x ) ^{2}} : \frac{- 2 c x ^{3} - 3 c ^{2} x ^{2} + c ^{4}}{2.4 ( c ^{2} + c x ) ^{2}}

= \frac{- 2 c x ^{3} - 3 c ^{2} x ^{2} + c ^{4}}{2.4 ( c ^{2} + c x ) ^{2}}

\frac{d y}{d x} = y (2 - y)

\int 2 x (x^{2} + 19)^{14} d x

\int (x^{2} - 5)^{5} 2 x d x

\int 7 arctan (x) d x

\frac{\partial}{\partial x} (\frac{2 ( y ^{2} - 4 )}{3 ( x - 1 ) ^{\frac{1}{3}}})

What is the derivative of (x(c^2-x^2)/(2.4(c^2+cx))) ?
The derivative of (x(c^2-x^2)/(2.4(c^2+cx))) is (-2cx^3-3c^2x^2+c^4)/(2.4(c^2+cx)^2)
What is the first derivative of (x(c^2-x^2)/(2.4(c^2+cx))) ?
The first derivative of (x(c^2-x^2)/(2.4(c^2+cx))) is (-2cx^3-3c^2x^2+c^4)/(2.4(c^2+cx)^2)