What is the derivative of sqrt((4+x^2)/(4-x)) ?

The derivative of sqrt((4+x^2)/(4-x)) is (-x^2+8x+4)/(2sqrt(x^2+4)(-x+4)^{3/2)}

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

The first derivative of sqrt((4+x^2)/(4-x)) is (-x^2+8x+4)/(2sqrt(x^2+4)(-x+4)^{3/2)}

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derivative of

\frac{4 + x ^{2}}{4 - x}

\frac{- x ^{2} + 8 x + 4}{2 x ^{2} + 4 ( - x + 4 ) ^{\frac{3}{2}}}

Solution steps

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

Apply the chain rule:

\frac{1}{2 \frac{4 + x ^{2}}{4 - x}} \frac{d}{d x} (\frac{4 + x ^{2}}{4 - x})

= \frac{1}{2 \frac{4 + x ^{2}}{4 - x}} \frac{d}{d x} (\frac{4 + x ^{2}}{4 - x})

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

= \frac{1}{2 \frac{4 + x ^{2}}{4 - x}} \cdot \frac{- x ^{2} + 8 x + 4}{( 4 - x ) ^{2}}

Simplify

\frac{1}{2 \frac{4 + x ^{2}}{4 - x}} \cdot \frac{- x ^{2} + 8 x + 4}{( 4 - x ) ^{2}} : \frac{- x ^{2} + 8 x + 4}{2 x ^{2} + 4 ( - x + 4 ) ^{\frac{3}{2}}}

= \frac{- x ^{2} + 8 x + 4}{2 x ^{2} + 4 ( - x + 4 ) ^{\frac{3}{2}}}

t an g e n t f (x) = (12 x - 8)^{\frac{1}{2}}, at x = 2

\frac{d y}{d x} = 7 x y^{2}

\frac{\partial}{\partial x} (3 x^{2} y + 2 y)

\int_{1}^{2} x e^{x} d x

d er i v a t i v e f (x) = 5^{s i n (x)}

What is the derivative of sqrt((4+x^2)/(4-x)) ?
The derivative of sqrt((4+x^2)/(4-x)) is (-x^2+8x+4)/(2sqrt(x^2+4)(-x+4)^{3/2)}
What is the first derivative of sqrt((4+x^2)/(4-x)) ?
The first derivative of sqrt((4+x^2)/(4-x)) is (-x^2+8x+4)/(2sqrt(x^2+4)(-x+4)^{3/2)}