*Notes to a video lecture on http://www.unizor.com*

__Derivative Example -__

Inverse Trigonometric Functions

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

**[**and its values are in

*−1,1*]**[**.

*−π/2,π/2*]Then, for any value of the argument

**from its domain the function value**

*x***is defined as an angle in radians such that**

*y*(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,

**and**

*A(x)=sin(arcsin(x))***, are equal within domain**

*B(x)=x***[**, their derivatives are equal as well.

*−1,1*]The derivative of the

**can be obtained using the chain rule for compounded functions.**

*A(x)*

*A*

^{I}**(x)=**d/dx**[**

*sin(arcsin(x))*]*=*

cos(arcsin(x))·cos(arcsin(x))·

*d/dx*

**[**

*arcsin(x)*]The derivative of

**is trivial.**

*B(x)*

*B**=*

^{I}**(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

**and**

*sin(arcsin(x))=x***∈**

*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***is not differentiable.**

*arcsin(x)*Geometrically, it signifies that tangential lines at both ends,

*x=−1***, of the domain of function**

*x=1***are vertical, as can be seen from a graph of this function below:**

*arcsin(x)*
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