Find the derivative of the function.
y = x arccos x - (sq. rt. (1 - x^2))
To find the derivative of the function y = x arccos(x) - √(1 - x^2), we can use the rules of differentiation. Let's break it down step by step:
Step 1: Apply the product rule
The product rule states that if we have a function in the form f(x)g(x), the derivative is given by f'(x)g(x) + f(x)g'(x).
In our case, f(x) = x and g(x) = arccos(x) - √(1 - x^2). Let's find their derivatives.
f'(x) = 1 (the derivative of x with respect to x)
g'(x) = (arccos(x))' - (√(1 - x^2))' (applying the subtraction rule)
Step 2: Find the derivatives of arccos(x) and √(1 - x^2)
To find the derivative of arccos(x), we use the chain rule. Let's denote arccos(x) as u(x):
u(x) = arccos(x)
u'(x) = -1/√(1 - x^2) (the derivative of arccos(x) with respect to x)
To find the derivative of √(1 - x^2), we use the power rule and chain rule. Let's denote √(1 - x^2) as v(x):
v(x) = √(1 - x^2)
v'(x) = (1/2)(1 - x^2)^(-1/2)(-2x) (the derivative of √(1 - x^2) with respect to x)
Step 3: Plug in the derivatives into the product rule
Now that we have the derivatives of f(x), g(x), u(x), and v(x), we can use the product rule:
y' = f'(x)g(x) + f(x)g'(x)
= (1)(arccos(x) - √(1 - x^2)) + (x)((-1/√(1 - x^2)) - (1/2)(1 - x^2)^(-1/2)(-2x))
Simplifying further:
y' = arccos(x) - √(1 - x^2) - (x/√(1 - x^2)) + x^2/√(1 - x^2)
Therefore, the derivative of the function y = x arccos(x) - √(1 - x^2) is y' = arccos(x) - √(1 - x^2) - (x/√(1 - x^2)) + x^2/√(1 - x^2).