What is the derivative of x/(4-sqrt(16-x^2)) ?

The derivative of x/(4-sqrt(16-x^2)) is 4/((sqrt(-x^2+16)-4)\sqrt{-x^2+16)}

What is the first derivative of x/(4-sqrt(16-x^2)) ?

The first derivative of x/(4-sqrt(16-x^2)) is 4/((sqrt(-x^2+16)-4)\sqrt{-x^2+16)}

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\frac{d}{d x} (\frac{x}{4 - 16 - x ^{2}})

\frac{4}{( - x ^{2} + 16 - 4 ) - x ^{2} + 16}

Solution steps

\frac{d}{d x} (\frac{x}{4 - 16 - x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d x}{d x} ( 4 - 16 - x ^{2} ) - \frac{d}{d x} ( 4 - 16 - x ^{2} ) x}{( 4 - 16 - x ^{2} ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (4 - 16 - x^{2}) = \frac{x}{- x ^{2} + 16}

= \frac{1 \cdot ( 4 - 16 - x ^{2} ) - \frac{x}{- x ^{2} + 16} x}{( 4 - 16 - x ^{2} ) ^{2}}

Simplify

\frac{1 \cdot ( 4 - 16 - x ^{2} ) - \frac{x}{- x ^{2} + 16} x}{( 4 - 16 - x ^{2} ) ^{2}} : \frac{4}{( - x ^{2} + 16 - 4 ) - x ^{2} + 16}

= \frac{4}{( - x ^{2} + 16 - 4 ) - x ^{2} + 16}

f (x) = 5 - 2 x^{2}, (- 5, - 45)

\int_{0}^{2} (1 + x)^{\frac{1}{3}} \cdot 2 d x

x y^{'} + (1 + x \cdot cot (x)) y = 0

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

y = 3 - x^{2}, y = 11 - 6 x

What is the derivative of x/(4-sqrt(16-x^2)) ?
The derivative of x/(4-sqrt(16-x^2)) is 4/((sqrt(-x^2+16)-4)\sqrt{-x^2+16)}
What is the first derivative of x/(4-sqrt(16-x^2)) ?
The first derivative of x/(4-sqrt(16-x^2)) is 4/((sqrt(-x^2+16)-4)\sqrt{-x^2+16)}