The derivative of the function f(x) = tan⁻¹(x) is given by:
f'(x) = 1 / (1 + x²)
To understand why this is the case, let’s look at the process of differentiation for the inverse tangent function. The function tan⁻¹(x) represents the angle whose tangent is x. When we differentiate this using implicit differentiation, we can consider that:
If y = tan⁻¹(x), then it follows that tan(y) = x.
Now, differentiating both sides with respect to x gives:
As per the chain rule:
sec²(y) * dy/dx = 1
Here, sec²(y) can be expressed as 1 + tan²(y), since sec²(y) = 1 + tan²(y).
Substituting tan(y) = x back into this results in:
dy/dx = 1 / (1 + tan²(y))
Then, since tan(y) = x, we replace tan²(y) with x², yielding:
dy/dx = 1 / (1 + x²)
This means that the derivative of the function f(x) = tan⁻¹(x) is 1 / (1 + x²), which is valid for all values of x. This function will approach zero as x becomes very large or very small, and it is defined for all real numbers.