### Video Transcript

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

In this question, we need to differentiate the inverse tan or arctan of π₯ over π with respect to π₯. We will begin by letting π¦ equal our expression inverse tan of π₯ over π. Taking tan or the tangent of both sides of this equation gives us tan π¦ is equal to π₯ over π. 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 π¦ gives us sec squared π¦ multiplied by dπ¦ by dπ₯. As π is a constant, the right-hand side differentiates to one over π. Dividing both sides of the equation by sec squared π¦ gives us dπ¦ by dπ₯ is equal to one over π multiplied by sec squared π¦.

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

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

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

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