Show that
$\frac{d}{d x}\left(2 \tan ^{-1} \theta\right)=\frac{d}{d \theta}\left[\tan ^{-1}\left(\frac{2 \theta}{1-\theta^2}\right)\right]$
Answer (1)
Recognize the trigonometric identity: tan − 1 ( 1 − θ 2 2 θ ) = 2 tan − 1 ( θ ) .
Differentiate the right-hand side with respect to θ : d θ d [ 2 tan − 1 ( θ )] = 1 + θ 2 2 .
Differentiate the left-hand side with respect to x using the chain rule: d x d [ 2 tan − 1 ( θ )] = 1 + θ 2 2 ⋅ d x d θ .
Conclude that the equality holds if we consider derivatives with respect to θ , i.e., d θ d ( 2 tan − 1 θ ) = d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] , and the result is 1 + θ 2 2 .
Explanation
Problem Analysis We are asked to show that d x d ( 2 tan − 1 θ ) = d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] . This problem involves differentiation and the chain rule.
Simplifying the RHS Let's first simplify the right-hand side (RHS). We know the trigonometric identity tan ( 2 θ ) = 1 − t a n 2 ( θ ) 2 t a n ( θ ) . Therefore, tan − 1 ( 1 − θ 2 2 θ ) = 2 tan − 1 ( θ ) .
Differentiating the RHS Now, we can rewrite the RHS as d θ d [ 2 tan − 1 ( θ ) ] . Taking the derivative with respect to θ , we get d θ d [ 2 tan − 1 ( θ ) ] = 2 ⋅ 1 + θ 2 1 = 1 + θ 2 2 .
Differentiating the LHS Now let's consider the left-hand side (LHS), d x d ( 2 tan − 1 θ ) . Since θ is a function of x , we apply the chain rule: d x d ( 2 tan − 1 θ ) = 2 ⋅ 1 + θ 2 1 ⋅ d x d θ = 1 + θ 2 2 d x d θ .
Comparing LHS and RHS From the previous steps, we have: RHS = 1 + θ 2 2 and LHS = 1 + θ 2 2 d x d θ . The original equation is d x d ( 2 tan − 1 θ ) = d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] . Substituting the derivatives we found, we need to show that 1 + θ 2 2 d x d θ = 1 + θ 2 2 . This is only true if d x d θ = 1 . However, the problem statement is slightly ambiguous. It seems that the problem intended to show that d θ d ( 2 tan − 1 θ ) = d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] .
Final Result Since d θ d ( 2 tan − 1 θ ) = 1 + θ 2 2 and d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] = 1 + θ 2 2 , we have shown that d θ d ( 2 tan − 1 θ ) = d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] .
Examples
In robotics, understanding inverse trigonometric functions and their derivatives is crucial for controlling joint angles. For example, if a robot arm's position is described by θ ( t ) , where t is time, then d t d θ represents the angular velocity of the joint. Manipulating expressions involving tan − 1 and its derivatives allows engineers to precisely control the robot's movements and ensure smooth operation.
