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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x)

(- 5, 5), (5, - 5)

(ln (3))^{^{'}}

\frac{\partial}{\partial x} (z + 1 - x e^{y} cos (z))

y = (arctan (3 x))^{2}

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