[Inverse Functions]
f(x)= 5cos(x)sin^(-1)(x)
f'(x)=?
To find the derivative of the given function f(x) = 5cos(x)sin^(-1)(x), we can apply the product rule and the chain rule.
Let's break down the expression step by step:
f(x) = 5cos(x)sin^(-1)(x)
First, let's differentiate the function 5cos(x) with respect to x. The derivative of cos(x) is -sin(x), so we have:
f1(x) = 5(-sin(x))sin^(-1)(x)
Next, we need to differentiate the function sin^(-1)(x) with respect to x. This requires the application of the chain rule. The derivative of sin^(-1)(x) with respect to x is given by 1/sqrt(1-x^2), so we have:
f2(x) = 5cos(x)(1/sqrt(1-x^2))
Now, let's multiply f1(x) and f2(x) together to get the derivative of the entire function:
f'(x) = f1(x) * f2(x)
= 5(-sin(x))sin^(-1)(x) * (cos(x))/sqrt(1-x^2)
= -5sin(x)sin^(-1)(x)cos(x) / sqrt(1-x^2)
Therefore, the derivative of f(x) = 5cos(x)sin^(-1)(x) is f'(x) = -5sin(x)sin^(-1)(x)cos(x) / sqrt(1-x^2).
Note: The derivative of sin^(-1)(x) is obtained by using the chain rule and the derivative of cos(x) is -sin(x).