Unizor - Derivatives Example - arcsin(x)

Unizor - Creative Minds through Art of Mathematics - Math4Teens Notes to a video lecture on http://www.unizor.com Derivative Example - Inverse Trigonometric Functions f(x) = arcsin(x) f I(x) = 1/√1−x² Proof We will use the method of implicit differentiation to obtain the formula for a derivative of this function. Let's start from a definition of function arcsin(x). The domain of this function is [−1,1] and its values are in [−π/2,π/2]. Then, for any value of the argument x from its domain the function value y is defined as an angle in radians such that (a) sin(y)=x (b) −π/2 ≤ y ≤ π/2 The first statement can be expressed as sin(arcsin(x))=x Since these two functions, A(x)=sin(arcsin(x)) and B(x)=x, are equal within domain [−1,1], their derivatives are equal as well. The derivative of the A(x) can be obtained using the chain rule for compounded functions. AI(x)=d/dx[sin(arcsin(x))] = cos(arcsin(x))·d/dx[arcsin(x)] The derivative of B(x) is trivial. BI(x) = d/dx[x] = 1 From equality of these two derivatives we conclude d/dx[arcsin(x)]=1/cos(arcsin(x)) Now let's analyze the expression cos(arcsin(x)). We know that sin(arcsin(x))=x and arcsin(x)∈[−π/2,π/2]. Therefore, cos(arcsin(x)) ≥ 0. Hence, cos(arcsin(x)) = √1−x² The final formula for a derivative is [arcsin(x)]I = 1/√1−x² A small detail remains when x = ±1, which results in zero denominator. These are the points where our function arcsin(x) is not differentiable. Geometrically, it signifies that tangential lines at both ends, x=−1 and x=1, of the domain of function arcsin(x) are vertical, as can be seen from a graph of this function below: