# Question Video: The Derivative of an Inverse Tangent Function Mathematics • Higher Education

Find d/d𝑥 tan⁻¹ (𝑥/𝑎), where 𝑎 ≠ 0.

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### Video Transcript

Find d by d𝑥 of inverse tan of 𝑥 over 𝑎, where 𝑎 is not equal to zero.

In this question, we need to differentiate the inverse tan or arctan of 𝑥 over 𝑎 with respect to 𝑥. We will begin by letting 𝑦 equal our expression inverse tan of 𝑥 over 𝑎. Taking tan or the tangent of both sides of this equation gives us tan 𝑦 is equal to 𝑥 over 𝑎. We will then differentiate both sides of this equation with respect to 𝑥.

We know that the derivative of tan 𝑥 is sec squared 𝑥. This means that using our knowledge of implicit differentiation, differentiating tan 𝑦 gives us sec squared 𝑦 multiplied by d𝑦 by d𝑥. As 𝑎 is a constant, the right-hand side differentiates to one over 𝑎. Dividing both sides of the equation by sec squared 𝑦 gives us d𝑦 by d𝑥 is equal to one over 𝑎 multiplied by sec squared 𝑦.

Whilst we do have an expression for d𝑦 by d𝑥, this is currently in terms of 𝑦. We need to go back to the point where tan 𝑦 was equal to 𝑥 over 𝑎. If we square both sides of this equation, we get tan squared 𝑦 is equal to 𝑥 squared over 𝑎 squared. One of our trigonometrical identities stated that tan squared 𝜃 plus one was equal to sec squared 𝜃. Subtracting one from both sides means that tan squared 𝜃 is equal to sec squared 𝜃 minus one. tan squared 𝑦 is, therefore, equal to sec squared 𝑦 minus one, which we know is equal to 𝑥 squared over 𝑎 squared.

We can multiply through by 𝑎 squared so that 𝑎 squared sec squared 𝑦 minus 𝑎 squared is equal to 𝑥 squared. Adding 𝑎 squared to both sides of this equation gives us 𝑎 squared sec squared 𝑦 is equal to 𝑎 squared plus 𝑥 squared.

The left-hand side is very similar to our expression for d𝑦 by d𝑥. However, we need 𝑎, not 𝑎 squared. We can, therefore, divide both sides of our equation by 𝑎 so that 𝑎 sec squared 𝑦 is equal to 𝑎 squared plus 𝑥 squared over 𝑎.

We want the reciprocal of this. And to work out the reciprocal of any fraction, we simply swap the numerator and denominator. d𝑦 by d𝑥 is, therefore, equal to 𝑎 over 𝑎 squared plus 𝑥 squared. This is the derivative of inverse tan of 𝑥 over 𝑎 with respect to 𝑥.