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

Find d/dπ‘₯ tan⁻¹ (π‘₯/π‘Ž), where π‘Ž β‰  0.

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Video Transcript

Find d by dπ‘₯ of inverse tan of π‘₯ over π‘Ž, where π‘Ž is not equal to zero.

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

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

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

We can multiply through by π‘Ž squared so that π‘Ž squared sec squared 𝑦 minus π‘Ž squared is equal to π‘₯ squared. Adding π‘Ž squared to both sides of this equation gives us π‘Ž squared sec squared 𝑦 is equal to π‘Ž squared plus π‘₯ squared.

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

We want the reciprocal of this. And to work out the reciprocal of any fraction, we simply swap the numerator and denominator. d𝑦 by dπ‘₯ is, therefore, equal to π‘Ž over π‘Ž squared plus π‘₯ squared. This is the derivative of inverse tan of π‘₯ over π‘Ž with respect to π‘₯.

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