y=cos^-1[(1-x^2)/(1+x^2)]
find dy/dx
To find dy/dx, we need to differentiate the given equation with respect to x. In this case, we have y = cos^(-1)((1-x^2)/(1+x^2)), also written as y = arccos((1-x^2)/(1+x^2)).
To differentiate arccos function, we'll use the chain rule.
Chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
For the given equation, let's assign u = (1 - x^2)/(1 + x^2). This means y = arccos(u), where u = (1 - x^2)/(1 + x^2).
We need to find du/dx, which will help us calculate dy/dx using the chain rule.
To find du/dx, we'll differentiate u = (1 - x^2)/(1 + x^2) with respect to x.
Let's use the quotient rule to differentiate u:
du/dx = [(d/dx)(1 - x^2)(1 + x^2) - (1 - x^2)(d/dx)(1 + x^2)] / (1 + x^2)^2
Expanding the terms and simplifying, we get:
du/dx = [(2x(1 + x^2) + 2x(1 - x^2)) - (1 - x^2)(2x)] / (1 + x^2)^2
Simplifying further, we have:
du/dx = (2x + 2x^3 + 2x - 2x^3 - 2x + 2x^3) / (1 + x^2)^2
du/dx = 4x / (1 + x^2)^2
Now that we have du/dx, we can find dy/dx using the chain rule.
dy/dx = f'(u) * du/dx
The derivative of arccos(u) with respect to u is -1/sqrt(1-u^2) according to the derivative of the inverse cosine function.
Substituting this into the expression, we get:
dy/dx = (-1/√(1 - u^2)) * (4x / (1 + x^2)^2)
Finally, substituting u back in as u = (1 - x^2)/(1 + x^2), we have:
dy/dx = (-4x / ((1 + x^2)^2 * √(1 - ((1 - x^2)/(1 + x^2))^2)))
Simplifying further, we get:
dy/dx = (-4x / ((1 + x^2)^2 * √(1 - (1 - x^2)^2/(1 + x^2)^2)))