### Video Transcript

Differentiate π¦ equals sec π₯ minus three csc π₯.

Here we are asked to find a derivative of the function π¦ with respect to the variable π₯. This can be denoted by π¦ dash, also known as π¦ prime, or dπ¦ by dπ₯. Note that the function π¦ is a sum of two functions, sec π₯ which we will call π¦ one and negative three csc π₯ which we will call π¦ two.

So using the property that the derivative of a sum of functions is the sum of the derivatives of the functions and the property that the derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function. We can rewrite the derivative of π¦ with respect to π₯ as the derivative of sec π₯ with respect to π₯ plus the derivative of negative three csc π₯ with respect to π₯ which equals the derivative of sec π₯ with respect to π₯ minus three times the derivative of csc π₯ with respect to π₯.

So now, we just need to find the derivatives of sec π₯ and csc π₯ with respect to π₯. First, letβs recall the definitions of sec π₯ and csc π₯. sec π₯ and csc π₯ are just shorthand notations for the reciprocal trigonometric functions, one of a cos π₯ and one of a sin π₯, respectively. Sometimes cosec π₯ is written exactly as it is spoken. There is another reciprocal trigonometric function, known as cot π₯. And itβs equal to one over tan π₯.

You can remember the shorthand notations for the three reciprocal trigonometric functions by noting that the third letter in the shorthand notation is the same as the first letter of the trigonometric function whose reciprocal is being taken. The derivatives of the three reciprocal trigonometric functions are well-known results. The derivative of sec π₯ is equal to the product sec π₯ tan π₯. The derivative of csc π₯ is equal to the product negative csc π₯ cot π₯. The derivative of cot π₯ is negative csc squared π₯, which is just another way of writing negative one over sin squared π₯.

We can either memorize these derivatives or work them out on the spot. Letβs see how we could work them out. Notice that all the reciprocal trigonometric functions are quotients in the variable π₯, i.e., of the form π’ over π£ where π’ and π£ are functions of π₯. So in order to differentiate them, we can use the quotient rule. In this rule, first we differentiate the function in the numerator and multiply it by the function in the denominator. We then subtract from this the product of the derivative of the function in the denominator with the function in the numerator.

We then divide the result by the square of the function in the denominator. Letβs use the quotient rule to differentiate sec π₯. Let π’ equal the constant function one and π£ equal the function cos π₯. Then the derivative of π’ is zero. And the derivative of π£ is negative sin π₯. Here, we have used the property that the derivative of a constant function is zero, as well as a standard well-known derivative. Substituting π’ and π£ and their derivatives into the quotient rule, we obtain that the derivative of sec π₯ is zero multiplied by cos π₯ minus one times negative sin π₯ all divided by cos squared π₯. This simplifies to sin π₯ over cos squared π₯ which we can rewrite as one over cos π₯ multiplied by sin π₯ over cos π₯.

Now remembering the definition of sec π₯ as well as the identity, sin π₯ over cos π₯ equals tan π₯, we obtain that the derivative of sec π₯ with respect to π₯ is the product sec π₯ tan π₯. In a similar way, we can use the quotient rule to work out the derivative of csc π₯. In this case, we let π’ equal the constant function one and π£ equal the function sin π₯. These functions differentiate to zero and cos π₯, respectively. Substituting π’ and π£ and their derivatives into the quotient rule, we obtain that the derivative of csc π₯ is zero multiplied by sin π₯ minus one multiplied by cos π₯ all divided by sin squared π₯ which simplifies to negative cos π₯ over sin squared π₯. This can be rewritten as negative one over sin π₯ multiplied by cos π₯ over sin π₯.

Now, note that cos π₯ over sin π₯ equals one divided by sin π₯ over cos π₯ which equals one over tan π₯. Here, we have used the identity sin π₯ over cos π₯ equals tan π₯. So we can rewrite the derivative of csc π₯ as negative one over sin π₯ multiplied by one over tan π₯. Now remembering the definitions of csc π₯ and cot π₯, the derivative of csc π₯ with respect to π₯ turns out to be the product negative csc π₯ cot π₯. As an exercise, try showing that the derivative of cot π₯ is negative csc squared π₯.

Having obtained the derivatives of sec π₯ and csc π₯, it just remains to use them to find the derivative of our original function. π¦ equals sec π₯ minus three csc π₯. We deduced that the derivative of π¦ with respect to π₯ is equal to the derivative of sec π₯ with respect to π₯ minus three times the derivative of csc π₯ with respect to π₯. So substituting the derivatives of sec π₯ and csc π₯ in, we obtain that the derivative of π¦ with respect to π₯ is equal to sec π₯ tan π₯ minus three times negative csc π₯ cot π₯ which simplifies to sec π₯ tan π₯ plus three csc π₯ cot π₯.

This completes the answer to the question. It is important to note that whilst we can work out the derivatives of the reciprocal trigonometric functions on the spot using the quotient rule, we must memorize the derivatives of sin π₯ and cos π₯ to answer questions of this form.