Show that if $a$ is a constant,
(a)
[tex]\frac{d}{dx}[\tan ^{-1}(\frac{x}{a})]=\frac{a}{a^2+x^2}[/tex]
(b)
[tex]\frac{d}{dx}[\sin ^{-1}(\frac{x}{a})]=\frac{1}{\sqrt{a^2+x^2}}[/tex]
Answer (1)
Use the chain rule to find the derivative of tan − 1 ( a x ) , resulting in a 2 + x 2 a .
Apply the chain rule to find the derivative of sin − 1 ( a x ) , which simplifies to a 2 − x 2 1 .
Assume 0"> a > 0 to simplify the expression for the derivative of sin − 1 ( a x ) .
The derivative of tan − 1 ( a x ) is a 2 + x 2 a and the derivative of sin − 1 ( a x ) is a 2 − x 2 1 .
Explanation
Problem Analysis We are asked to prove the derivatives of two inverse trigonometric functions. Part (a) involves the derivative of tan − 1 ( a x ) , and part (b) involves the derivative of sin − 1 ( a x ) , where a is a constant.
Derivative of arctan(x/a) For part (a), we need to find d x d [ tan − 1 ( a x ) ] . We will use the chain rule. Let u = a x . Then, d x d u = a 1 . We know that d u d tan − 1 ( u ) = 1 + u 2 1 . Therefore, by the chain rule, we have d x d [ tan − 1 ( a x ) ] = 1 + ( a x ) 2 1 ⋅ a 1 = 1 + a 2 x 2 1 ⋅ a 1 = a 2 a 2 + x 2 1 ⋅ a 1 = a 2 + x 2 a 2 ⋅ a 1 = a 2 + x 2 a .
Derivative of arcsin(x/a) For part (b), we need to find d x d [ sin − 1 ( a x ) ] . We will again use the chain rule. Let u = a x . Then, d x d u = a 1 . We know that d u d sin − 1 ( u ) = 1 − u 2 1 . Therefore, by the chain rule, we have d x d [ sin − 1 ( a x ) ] = 1 − ( a x ) 2 1 ⋅ a 1 = 1 − a 2 x 2 1 ⋅ a 1 = a 2 a 2 − x 2 1 ⋅ a 1 = ∣ a ∣ a 2 − x 2 1 ⋅ a 1 = a 2 − x 2 ∣ a ∣ ⋅ a 1 .
Simplifying the Result Assuming 0"> a > 0 , we have ∣ a ∣ = a . Thus, d x d [ sin − 1 ( a x ) ] = a 2 − x 2 a ⋅ a 1 = a 2 − x 2 1 . Note that the original problem statement had a typo; it should be a 2 − x 2 1 instead of a 2 + x 2 1 .
Final Answer (a) The derivative of tan − 1 ( a x ) is a 2 + x 2 a .
(b) The derivative of sin − 1 ( a x ) is a 2 − x 2 1 .
Examples
Inverse trigonometric functions and their derivatives are essential in physics and engineering. For example, when analyzing the motion of a pendulum, the angle the pendulum makes with the vertical can be described using inverse trigonometric functions. The derivatives of these functions are then used to determine the pendulum's angular velocity and acceleration, which are crucial for understanding its dynamics.
