Find the derivative of the following inverse function: [tex]$y=x \sin ^{-1} x$[/tex]
Answer (1)
Apply the product rule: d x d ( uv ) = u ′ v + u v ′ , where u = x and v = sin − 1 x .
Find the derivatives: d x d ( x ) = 1 and d x d ( sin − 1 x ) = 1 − x 2 1 .
Substitute into the product rule: d x d y = ( 1 ) ( sin − 1 x ) + ( x ) ( 1 − x 2 1 ) .
Simplify the expression: d x d y = sin − 1 x + 1 − x 2 x .
Explanation
Problem Analysis We are given the function y = x sin − 1 x and we need to find its derivative with respect to x . This problem involves finding the derivative of a product of two functions, so we will use the product rule.
Applying the Product Rule The product rule states that if y = u ( x ) v ( x ) , then d x d y = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) . In our case, let u ( x ) = x and v ( x ) = sin − 1 x .
Derivative of u(x) First, we find the derivative of u ( x ) = x with respect to x :
d x d u = d x d ( x ) = 1
Derivative of v(x) Next, we find the derivative of v ( x ) = sin − 1 x with respect to x . Recall that the derivative of sin − 1 x is 1 − x 2 1 :
d x d v = d x d ( sin − 1 x ) = 1 − x 2 1
Applying the Product Rule Formula Now, we apply the product rule: d x d y = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) = 1 ⋅ sin − 1 x + x ⋅ 1 − x 2 1 So, we have: d x d y = sin − 1 x + 1 − x 2 x
Final Answer Therefore, the derivative of y = x sin − 1 x with respect to x is: d x d y = sin − 1 x + 1 − x 2 x
Examples
Understanding derivatives of inverse trigonometric functions is crucial in physics, especially when dealing with oscillatory motions or wave phenomena. For instance, when analyzing the motion of a pendulum, the angle can be expressed using inverse trigonometric functions, and finding the rate of change of this angle (its derivative) helps determine the pendulum's angular velocity. This is also applicable in electrical engineering when analyzing AC circuits involving inductors and capacitors, where phase angles are often described using inverse trigonometric functions.
