Derivative of Inverse Trigonometric Functions
f(x) = sin(arccos(4x))
What is f'(x)?
f(x)= sin(arccos(4x)) = sqrt(1 - 16x^2)
(To prove that, draw yourself a triangle with cos A = 4x and figure out the sin of A)
let u = 1 - 16x^2
f(u) = sqrt u
f'(x) = df/dx = df/du du/dx
= (1/2)(u)^-1/2 * -32 x
= -16 x/sqrt[1 - 16x^2]
Thanks. The next question I have is pretty similar.
f(x) =cos(arcsin(2x))
Find f'(x)
It is so similar that you should be able to do it the same way. Draw the triangle and you have sqrt(1-4x^2)
f(x) = (1-4x^2)^.5
f'(x) = .5 [(1-4x^2)^-.5] (-8x)
= -4x/sqrt(1-4x^2)
To find the derivative of the function f(x) = sin(arccos(4x)), we can use the chain rule. The chain rule states that if we have a composition of two functions, g(f(x)), then the derivative of g(f(x)) with respect to x is given by g'(f(x)) multiplied by f'(x).
Let's break down the function f(x):
f(x) = sin(arccos(4x))
We can see that f(x) is a composition of two functions: sin(x) and arccos(4x).
First, let's find the derivative of the inner function arccos(4x):
The derivative of arccos(u) with respect to u is -1/sqrt(1 - u^2), where u is a function of x.
In this case, u = 4x. So, the derivative of arccos(4x) with respect to x is:
d(arccos(4x))/dx = -1/sqrt(1 - (4x)^2) = -1/sqrt(1 - 16x^2)
Next, let's find the derivative of the outer function sin(x):
The derivative of sin(x) with respect to x is cos(x).
Now, using the chain rule, we can find the derivative of f(x):
f'(x) = cos(arccos(4x)) * (-1/sqrt(1 - 16x^2))
Notice that cos(arccos(4x)) simplifies to just 4x, since cos(arccos(u)) is equal to u.
Therefore, f'(x) = 4x * (-1/sqrt(1 - 16x^2)) = -4x/sqrt(1 - 16x^2)
So, the derivative of f(x) = sin(arccos(4x)) is f'(x) = -4x/sqrt(1 - 16x^2).