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 (2)
We demonstrated that the derivatives of both sides of the equation are equal under the condition that d x d θ = 1 . This implies that θ changes linearly with respect to x .
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Evaluate the derivative of the RHS, d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] , using the chain rule and quotient rule, simplifying to 1 + θ 2 2 .
Evaluate the derivative of the LHS, d x d ( 2 tan − 1 θ ) , using the chain rule, resulting in 1 + θ 2 2 d x d θ .
Equate the LHS and RHS derivatives: 1 + θ 2 2 d x d θ = 1 + θ 2 2 .
Deduce that the given equation holds if d x d θ = 1 .
Explanation
Problem Setup We are asked to show that d x d ( 2 tan − 1 θ ) = d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] where θ is a function of x , i.e., θ ( x ) .
Evaluating the RHS Derivative Let's first evaluate the derivative on the right-hand side (RHS) with respect to θ . We have d θ d [ tan − 1 ( 1 − θ 2 2 θ ) ] Recall that d x d tan − 1 ( x ) = 1 + x 2 1 . Applying the chain rule, we get d θ d tan − 1 ( 1 − θ 2 2 θ ) = 1 + ( 1 − θ 2 2 θ ) 2 1 ⋅ d θ d ( 1 − θ 2 2 θ ) Now, we need to find the derivative of 1 − θ 2 2 θ with respect to θ . Using the quotient rule, we have d θ d ( 1 − θ 2 2 θ ) = ( 1 − θ 2 ) 2 ( 1 − θ 2 ) ( 2 ) − ( 2 θ ) ( − 2 θ ) = ( 1 − θ 2 ) 2 2 − 2 θ 2 + 4 θ 2 = ( 1 − θ 2 ) 2 2 + 2 θ 2 = ( 1 − θ 2 ) 2 2 ( 1 + θ 2 ) Substituting this back into the expression for the derivative of the RHS, we get 1 + ( 1 − θ 2 ) 2 4 θ 2 1 ⋅ ( 1 − θ 2 ) 2 2 ( 1 + θ 2 ) = ( 1 − θ 2 ) 2 + 4 θ 2 ( 1 − θ 2 ) 2 ⋅ ( 1 − θ 2 ) 2 2 ( 1 + θ 2 ) = ( 1 − θ 2 ) 2 + 4 θ 2 2 ( 1 + θ 2 ) Simplifying the denominator, we have ( 1 − θ 2 ) 2 + 4 θ 2 = 1 − 2 θ 2 + θ 4 + 4 θ 2 = 1 + 2 θ 2 + θ 4 = ( 1 + θ 2 ) 2 . Therefore, ( 1 + θ 2 ) 2 2 ( 1 + θ 2 ) = 1 + θ 2 2 So, the derivative of the RHS with respect to θ is 1 + θ 2 2 .
Evaluating the LHS Derivative Now, let's evaluate the derivative on the left-hand side (LHS) with respect to x . We have 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 So, the derivative of the LHS with respect to x is 1 + θ 2 2 d x d θ .
Equating LHS and RHS Now, we want to show that the LHS is equal to the RHS, i.e., 1 + θ 2 2 d x d θ = 1 + θ 2 2 Dividing both sides by 1 + θ 2 2 , we get d x d θ = 1 This implies that θ = x + C , where C is a constant. Therefore, the given equation holds if d x d θ = 1 .
Final Condition The given equation holds if d x d θ = 1 . This means that θ must be equal to x + C , where C is a constant.
Examples
In control systems, understanding how the derivative of an angle (represented by θ ) changes with respect to time ( x ) is crucial for designing stable feedback loops. If the rate of change of the angle with respect to time is constant (i.e., d x d θ = 1 ), it simplifies the design and analysis of the control system, ensuring predictable and stable behavior. This concept is used in robotics, aerospace, and process control to maintain desired orientations or positions.
