for abs value of x <1, the derivative of y= ln square root 1-x^2 is.....?
To find the derivative of the given function, y = ln √(1-x^2), we can use the chain rule. The chain rule states that if we have a composition of functions, the derivative can be found by taking the derivative of the outer function multiplied by the derivative of the inner function.
Let's start by breaking down the function:
y = ln √(1-x^2)
We can rewrite √(1-x^2) as (1-x^2)^(1/2):
y = ln [(1-x^2)^(1/2)]
Now, let's define a new function u = (1-x^2)^(1/2). The function y can be written as:
y = ln u
The derivative of y with respect to x can be written as:
dy/dx = (dy/du) * (du/dx)
Now, let's find these derivatives individually:
1. First, we find (dy/du):
Since y = ln u, the derivative of y with respect to u is simply 1/u. So, (dy/du) = 1/u.
2. Next, we find (du/dx):
By using the power rule, we can find the derivative of u = (1-x^2)^(1/2) with respect to x as follows:
(du/dx) = (1/2)(1-x^2)^(-1/2)(-2x)
= -x / (1-x^2)^(1/2)
Now, we can multiply these two derivatives together to find the overall derivative:
dy/dx = (dy/du) * (du/dx)
= (1/u) * (-x / (1-x^2)^(1/2))
= -x / [u(1-x^2)^(1/2)]
Since u = (1-x^2)^(1/2), we can substitute it back into the equation:
dy/dx = -x / [(1-x^2)^(1/2) * (1-x^2)^(1/2)]
= -x / (1-x^2)
Therefore, the derivative of y = ln √(1-x^2) is dy/dx = -x / (1-x^2).
Given the condition that |x| < 1, this derivative is valid for the given interval.