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))

\int_{- 1}^{1} 9 x^{2} + 6 x - 3 d x

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

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

(- x^{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))