Derivative of arccos(x)
To find the derivative of the inverse trigonometric function, arccos(x), we can use the chain rule
To find the derivative of the inverse trigonometric function, arccos(x), we can use the chain rule. Recall that the chain rule states that if we have a composition of functions, the derivative of the whole composition is the derivative of the outer function multiplied by the derivative of the inner function.
Let’s begin by noting that arccos(x) is the inverse function of cosine(x). So we can rewrite it as:
y = arccos(x)
x = cos(y)
Taking the derivative of both sides with respect to x, we get:
1 = -sin(y) * dy/dx
Next, we need to find dy/dx, since we want to find d(arccos(x))/dx.
Rearranging the equation, we have:
dy/dx = 1/(-sin(y))
Now, recall the Pythagorean identity:
sin^2(y) + cos^2(y) = 1
Since y is equivalent to arccos(x), we can substitute x for cos(y):
sin^2(y) + x^2 = 1
Solving for sin(y), we have:
sin(y) = sqrt(1 – x^2)
Substituting this expression back into the equation for dy/dx, we find:
dy/dx = 1/(-sqrt(1 – x^2))
Therefore, the derivative of arccos(x) is:
d(arccos(x))/dx = 1/(-sqrt(1 – x^2))
This is the detailed answer for the derivative of arccos(x) using the chain rule.
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