Differentiate.
cos^-1 2x
is this arccos (2x)?
or do you mean 1/(cos 2x)
I think we can assume arccos.
If so, then recall that d/dx (arccos(u)) = -1/√(1-u^2) du/dx
Now just plug and chug.
To differentiate the function cos^-1(2x), we can use the chain rule.
The chain rule states that if we have a composite function, f(g(x)), then the derivative of f(g(x)) is given by f'(g(x)) * g'(x).
Let's break down the function cos^-1(2x) into its component parts:
f(u) = cos^-1(u)
g(x) = 2x
Now, let's find the derivatives of f(u) and g(x):
f'(u) = -1/√(1-u^2) (derivative of cos^-1(u))
g'(x) = 2 (derivative of 2x)
Using the chain rule, the derivative of cos^-1(2x) is given by:
f'(g(x)) * g'(x) = [-1/√(1-(2x)^2)] * 2
Simplifying further:
= -2/√(1-4x^2)
Therefore, the derivative of cos^-1(2x) is -2/√(1-4x^2).