# Question Video: Finding the Second Derivative of a Function Defined by Parametric Equations Involving Trigonometric Functions Mathematics • Higher Education

If π₯ = 2 sec 5π§ and β(3π¦) = tan 5π§, find dΒ²π¦/dπ₯Β².

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### Video Transcript

If π₯ is equal to two multiplied by the sec of five π§ and the square root of three π¦ is equal to the tan of five π§, find the second derivative of π¦ with respect to π₯.

Since the question asks us to find the second derivative of π¦ with respect to π₯, it might be tempting to try and write π¦ in terms of π₯. However, thereβs a simpler method by using the chain rule. We recall that calculating the second derivative of π¦ with respect to π₯ is the same as differentiating dπ¦ dπ₯ with respect to π₯. And we recall that the chain rule tells us that the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π§ multiplied by the derivative of π§ with respect to π₯. And instead of multiplying by the derivative of π§ with respect to π₯, since weβre given π₯ as a function of π§, we can do the equivalent, dividing by the derivative of π₯ with respect to π§.

So letβs start by calculating the derivative of π₯ with respect to π§. Thatβs equal to the derivative of two multiplied by the sec of five π§ with respect to π§. And we recall that the derivative of the sec of ππ₯ with respect to π₯ is equal to π multiplied by the sec of ππ₯ multiplied by the tan of ππ₯. Applying this gives us that the derivative of π₯ with respect to π§ is equal to two multiplied by five, which we can simplify to 10, multiplied by the sec of five π§ multiplied by the tan of five π§.

Calculating the derivative of π¦ with respect to π§ is more difficult since weβre not given π¦ directly as a function of π§. The simplest way to do this is to rearrange our equation to make π¦ the subject. Weβll do this by squaring both sides of the equation. Evaluating this gives us that three π¦ is equal to the tan squared of five π§.

Next, we differentiate both sides of our equation with respect to π§. Differentiating three π¦ with respect to π§ gives us three multiplied by the derivative of π¦ with respect to π§. We can calculate the derivative of the tan of five π§ all squared with respect to π§. By recalling that, for a constant π, the derivative with respect to π₯ of the tan of π multiplied by π₯ is equal to π multiplied by the sec of ππ₯ squared.

And by using the chain rule, we have that the derivative of the function π of π₯ all squared is equal to two multiplied by π prime of π₯ multiplied by π of π₯. Using this gives us that the derivative of the tan of five π§ all squared with respect to π§ is equal to two multiplied by our function π prime of π§. Which weβve shown to be equal to five multiplied by the sec of five π§ all squared. And then we multiply this by our function π of π§, which is the tan of five π§.

We can simplify this by multiplying the two and the five to get 10 and then dividing both sides of our equation by three. This gives us that the derivative of π¦ with respect to π§ is equal to 10 over three multiplied by the sec of five π§ squared multiplied by the tan of five π§.

Weβre now ready to apply our chain rule that the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π§ divided by the derivative of π₯ with respect to π§. We have a numerator of dπ¦ dπ§, which is equal to ten-thirds multiplied by the sec of five π§ all squared multiplied by the tan of five π§. And a denominator of dπ₯ by dπ§, which is equal to 10 multiplied by the sec of five π§ multiplied by the tan of five π§.

We can then start canceling the shared factors. We can cancel the shared factor of the tan of five π§ in the numerator and the denominator. We can cancel one shared factor of the sec of five π§ in the numerator and the denominator. And we can cancel the shared factor of 10 in the numerator and the denominator.

Therefore, weβve shown that the derivative of π¦ with respect to π₯ is equal to the sec of five π§ all divided by three. So weβve shown that our derivative function dπ¦ dπ₯ is a function of π§. Letβs call this π of π§. And the question wants us to calculate the second derivative of π¦ with respect to π₯. And thatβs the same as differentiating our function π of π§ with respect to π₯.

And weβve already seen how to do this. We can use the chain rule. This gives us the derivative of π with respect to π₯ is equal to the derivative of π with respect to π§ multiplied by the derivative of π§ with respect to π₯. And just as we did before, instead of multiplying by the derivative of π§ with respect to π₯, since weβre given π₯ as a function of π§, we can instead divide by the derivative of π₯ with respect to π§.

So what we have is the second derivative of π¦ with respect to π₯ is equal to the derivative of dπ¦ dπ₯ with respect to π₯. And weβve called dπ¦ dπ₯ π of π§. And then using the chain rule, weβve shown that this is equal to dπ dπ§ divided by dπ₯ dπ§. Substituting in our expression for π, we get the derivative of π with respect to π§ is equal to the derivative of the sec of five π§ all divided by three with respect to π§.

Then we can substitute in our expression for dπ₯ by dπ§. This is 10 sec of five π§ multiplied by the tan of five π§. We can differentiate the sec of five π§ to be equal to five multiplied by the sec of five π§ multiplied by the tan of five π§. And then we divide through by our constant of three. And instead of dividing, weβre going to multiply by the reciprocal.

Simplifying this gives us five multiplied by the sec of five π§ multiplied by the tan of five π§ all divided by 30 multiplied by the sec of five π§ multiplied by the tan of five π§. We can then cancel the shared factor of the tan of five π§ in our numerator and our denominator and the shared factor of the sec of five π§ in the numerator and the denominator. And if we then cancel the shared factor of five, we get one-sixth.

So weβve shown that if π₯ is equal to two multiplied by the sec of five π§ and the square root of three π¦ is equal to the tan of five π§, then the second derivative of π¦ with respect to π₯ is equal to one-sixth.