### Video Transcript

Given π¦ equals sin four π₯ tan four π₯, find dπ¦ by dπ₯ at π₯ equals π by six.

In this question, weβre asked to find dπ¦ by dπ₯, thatβs the first derivative of this function π¦, and then evaluate it at a particular π₯-value, the π₯-value π by six. We can see that the function weβve been given is the product of two functions. Itβs equal to sin four π₯ multiplied by tan four π₯. And so, in order to find its derivative, weβre going to need to apply the product rule.

This tells us that for two differentiable functions π’ and π£, the derivative with respect to π₯ of their product π’π£ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. We multiply each function by the derivative of the other and add them together. We can, therefore, define π’ to be the function sin four π₯ and π£ to be the second factor, π£ is equal to tan four π₯. We then need to find each of their individual derivatives. And to do so, weβre going to need to recall some standard results for differentiation of trigonometric functions.

Firstly, we recall that the derivative with respect to π₯ of sin of ππ₯ for some constant π is equal to π multiplied by cos of ππ₯. And this can be proven from first principles. Although we must remember that these results for trigonometric functions are only valid if the angle is measured in radians. Secondly, we recall that the derivative with respect to π₯ of tan ππ₯ is equal to π multiplied by sec squared ππ₯. And we can prove this by recalling that tan of ππ₯ is equal to sin ππ₯ over cos ππ₯. We can then recall the standard results for differentiating sin of ππ₯ and cos of ππ₯ and apply the quotient rule.

In each case here, the value of the constant π is four. So, we have that the derivative of sin four π₯ with respect to π₯ is four cos four π₯ and that the derivative of tan four π₯ with respect to π₯ is four sec squared four π₯. Now, weβre ready to substitute into the product. Well, we have that dπ¦ by dπ₯ is equal to π’ times dπ£ by dπ₯, thatβs sin four π₯ multiplied by four sec squared four π₯, plus π£ times dπ’ by dπ₯, thatβs tan four π₯ multiplied by four cos four π₯.

We can then manipulate this expression slightly. In the first term, we have sec squared four π₯, which is equal to one over cos squared four π₯. So, we can write the first term as four sin four π₯ over cos four π₯ multiplied by one over cos four π₯. In the second term, we have tan four π₯, which can be written as sin of four π₯ over cos four π₯. So, the second term becomes four sin four π₯ over cos four π₯ multiplied by cos four π₯.

Within the second term then, we see that a factor of cos four π₯ can be cancelled from both the numerator and denominator to leave the simplified term four sin four π₯. And in the first term, where we have sin four π₯ over cos four π₯, we could rewrite this as tan four π₯. So, our simplified expression for dπ¦ by dπ₯ is four tan four π₯ over cos four π₯ plus four sin four π₯.

Next, we need to evaluate this expression when π₯ is equal to π by six. Four π₯ will, therefore, be equal to four π by six, which simplifies to two π by three. When π₯ is equal to π by six then, dπ¦ by dπ₯ will be equal to four tan two π by three over cos of two π by three plus four sin two π by three. Now, if we have a calculator, we can evaluate this directly. But if not, we can still answer the question. Because two π by three is one of those special angles for which we need to know the values of its trigonometric ratios of by heart.

We should recall that tan of two π by three is equal to negative root three and cos of two π by three is equal to negative one-half. In the same way, sin of two π by three is equal to root three over two. So, we have exact values for each of these trigonometric ratios. We, therefore, have negative four root three over negative a half, which is the same as negative four root three multiplied by negative two, plus four root three over two.

The first term here will simplify to eight root three. And in the second term, we can cancel a common factor of two to leave us with two root three. So, we have that dπ¦ by dπ₯, when π₯ equals π by six, is equal to eight root three plus two root three, which is of course 10 root three. So, weβve applied the product rule and standard results for differentiation of trigonometric functions to find that if π¦ is the function sin four π₯ multiplied by tan four π₯, then its first derivative dπ¦ by dπ₯ at π₯ equals π by six is 10 root three.