The Derivative of f(x) = (tanx)^-1 Using the Chain Rule for Differentiation

d/dx[(tanx)^-1]

To find the derivative of the function f(x) = (tanx)^-1, where ‘(d/dx)’ represents derivative with respect to x, we use the chain rule of differentiation

To find the derivative of the function f(x) = (tanx)^-1, where ‘(d/dx)’ represents derivative with respect to x, we use the chain rule of differentiation.

First, let’s rewrite the function f(x) using an exponent instead of the negative power:

f(x) = (tanx)^-1 = 1/(tanx)

Now, let u = tanx. We can rewrite f(x) as:

f(x) = 1/u

To find the derivative of f(x) with respect to x, we can use the chain rule, which states that if y = u^n, then dy/dx = n*u^(n-1) * du/dx.

Applying the chain rule to our function, we have:

f'(x) = d/dx[1/u]
= d/dx[tanx]^-1
= -1/(tanx)^2 * d/dx[tanx]

The derivative of tanx with respect to x is calculated using the derivative of the tan function, which is sec^2x:

d/dx[tanx] = sec^2x

Substituting this back into our equation, we have:

f'(x) = -1/(tanx)^2 * sec^2x

However, this can still be further simplified. Recall that sec^2x is equal to 1 + tan^2x. By substituting this in, we can simplify the expression further:

f'(x) = -1/(tanx)^2 * sec^2x
= -1/(tanx)^2 * (1 + tan^2x)
= -1/(tanx)^2 * (1 + (sinx/cosx)^2)
= -1/(tanx)^2 * (1 + sin^2x/cos^2x)
= -1/(tanx)^2 * (cos^2x + sin^2x)/cos^2x
= -1/(tanx)^2 * 1/cos^2x
= -cos^2x/(sin^2x * cos^2x)
= -cos^2x/sin^2x * 1/cos^2x
= -1/sin^2x

Thus, the derivative of f(x) = (tanx)^-1 with respect to x is -1/sin^2x.

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