What is the derivative of f(x)=(4-x^2)/(8+x^2) ?

The derivative of f(x)=(4-x^2)/(8+x^2) is -(24x)/((8+x^2)^2)

What is the first derivative of f(x)=(4-x^2)/(8+x^2) ?

The first derivative of f(x)=(4-x^2)/(8+x^2) is -(24x)/((8+x^2)^2)

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derivative of

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

- \frac{24 x}{( 8 + x ^{2} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( - 2 x ) ( 8 + x ^{2} ) - 2 x ( 4 - x ^{2} )}{( 8 + x ^{2} ) ^{2}} : - \frac{24 x}{( 8 + x ^{2} ) ^{2}}

= - \frac{24 x}{( 8 + x ^{2} ) ^{2}}

\int 2 x (6 x^{2} - 12)^{2} d x

\frac{\partial}{\partial x} (e^{x y} + 5 x - 7 y + 3)

x \to 0 - lim (\frac{- 3 x ^{3} + 4 x}{x})

y^{^{''}} + 2 y^{^{'}} - 8 y = 6 e^{- 2 x} - e^{- x}

\frac{d}{d x} (\frac{1 + sin ( x )}{1 - sin ( x )})

What is the derivative of f(x)=(4-x^2)/(8+x^2) ?
The derivative of f(x)=(4-x^2)/(8+x^2) is -(24x)/((8+x^2)^2)
What is the first derivative of f(x)=(4-x^2)/(8+x^2) ?
The first derivative of f(x)=(4-x^2)/(8+x^2) is -(24x)/((8+x^2)^2)