What is the derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 ?

The derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 is (x^2)/(sqrt(-x^2+16))

What is the first derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 ?

The first derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 is (x^2)/(sqrt(-x^2+16))

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derivative of

y = 8 arcsin (\frac{x}{4}) - \frac{x 16 - x ^{2}}{2}

\frac{x ^{2}}{- x ^{2} + 16}

Solution steps

\frac{d}{d x} (8 arcsin (\frac{x}{4}) - \frac{x 16 - x ^{2}}{2})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (8 arcsin (\frac{x}{4})) - \frac{d}{d x} (\frac{x 16 - x ^{2}}{2})

\frac{d}{d x} (8 arcsin (\frac{x}{4})) = \frac{8}{16 - x ^{2}}

\frac{d}{d x} (\frac{x 16 - x ^{2}}{2}) = \frac{8 - x ^{2}}{16 - x ^{2}}

= \frac{8}{16 - x ^{2}} - \frac{8 - x ^{2}}{16 - x ^{2}}

Simplify

\frac{8}{16 - x ^{2}} - \frac{8 - x ^{2}}{16 - x ^{2}} : \frac{x ^{2}}{- x ^{2} + 16}

= \frac{x ^{2}}{- x ^{2} + 16}

\frac{\partial}{\partial x} (e^{y} cos (x))

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

l a pl a ce t r an s f or m e^{- 2 t} sin (t)

x \to a lim (\frac{3}{0})

d er i v a t i v e \frac{- x ^{2} - 4}{( x ^{2} - 4 ) ^{2}}

What is the derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 ?
The derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 is (x^2)/(sqrt(-x^2+16))
What is the first derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 ?
The first derivative of y=8arcsin(x/4)-(xsqrt(16-x^2))/2 is (x^2)/(sqrt(-x^2+16))