What is the derivative of x^9arccos(x) ?

The derivative of x^9arccos(x) is 9x^8arccos(x)-(x^9)/(sqrt(1-x^2))

What is the first derivative of x^9arccos(x) ?

The first derivative of x^9arccos(x) is 9x^8arccos(x)-(x^9)/(sqrt(1-x^2))

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derivative of

x^{9} arccos (x)

9 x^{8} arccos (x) - \frac{x ^{9}}{1 - x ^{2}}

Solution steps

\frac{d}{d x} (x^{9} arccos (x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{9}, g = arccos (x)

= \frac{d}{d x} (x^{9}) arccos (x) + \frac{d}{d x} (arccos (x)) x^{9}

\frac{d}{d x} (x^{9}) = 9 x^{8}

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

= 9 x^{8} arccos (x) + (- \frac{1}{1 - x ^{2}}) x^{9}

Simplify

9 x^{8} arccos (x) + (- \frac{1}{1 - x ^{2}}) x^{9} : 9 x^{8} arccos (x) - \frac{x ^{9}}{1 - x ^{2}}

= 9 x^{8} arccos (x) - \frac{x ^{9}}{1 - x ^{2}}

\int \frac{x}{e ^{3 x}} d x

\int ln (x) x^{- 3} d x

\frac{\partial}{\partial x} (x^{2} - 6 y)

(- ln (2 x))^{^{'}}

f (x) = x^{2} + 1, at x = 2

What is the derivative of x^9arccos(x) ?
The derivative of x^9arccos(x) is 9x^8arccos(x)-(x^9)/(sqrt(1-x^2))
What is the first derivative of x^9arccos(x) ?
The first derivative of x^9arccos(x) is 9x^8arccos(x)-(x^9)/(sqrt(1-x^2))