Show that the derivative of y= cosh-1 (2x-1) is:
dy/dx = √1/(x2-x)
well, we know that d/dx arccosh u = 1/√(u^2-1) du
So just use the chain rule, and note that
(2x-1)^2-1 = 4x^2-4x
To find the derivative of y = cosh^(-1)(2x-1), we can use the chain rule. The chain rule states that if we have a composite function y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = f'(g(x)) * g'(x), where f'(g(x)) represents the derivative of the outer function f evaluated at g(x), and g'(x) represents the derivative of the inner function g.
Let's break down the given function step by step to apply the chain rule.
Step 1: Identify the outer function f(u) and inner function g(x):
In the given function, the outer function is f(u) = cosh^(-1)(u), and the inner function is g(x) = 2x-1. Here, u is the variable inside the inverse hyperbolic cosine function.
Step 2: Find the derivatives of the outer and inner function:
The derivative of the outer function f(u) = cosh^(-1)(u) can be found by taking the derivative of the inverse hyperbolic cosine function, which is given by d(u) / du = 1 / √(u^2 - 1).
The derivative of the inner function g(x) = 2x-1 is simply g'(x) = 2.
Step 3: Apply the chain rule:
Now we can apply the chain rule to find the derivative of the composite function y = cosh^(-1)(2x-1):
dy/dx = f'(g(x)) * g'(x)
= [1 / √((2x-1)^2 - 1)] * 2
= 2 / √((2x-1)^2 - 1)
Simplifying the expression gives:
dy/dx = √(1 / ((2x-1)^2 - 1))
= √(1 / (4x^2 - 4x + 1 - 1))
= √(1 / (4x^2 - 4x))
= √(1 / (4x(x - 1)))
= √(1 / (x(x - 1)))
= √(1 / (x^2 - x))
Therefore, the derivative of y = cosh^(-1)(2x-1) is dy/dx = √(1 / (x^2 - x)).