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Question Video: Evaluating Higher Order Derivatives of a Combination of Polynomial and Trigonometric Functions Mathematics • Higher Education

Evaluate d/dπ‘₯ [βˆ’3π‘₯Β³ + (d/dπ‘₯)(2π‘₯⁡ βˆ’ 9 sec π‘₯)].

03:24

Video Transcript

Evaluate d by dπ‘₯ of negative three π‘₯ cubed plus d by dπ‘₯ of two π‘₯ to the fifth power minus nine times the sec of π‘₯.

We’re asked to evaluate an expression, and we can see this expression contains the derivative with respect to π‘₯ twice. There’s a few different ways of doing this. However, the easiest way is to evaluate the innermost derivatives first. So we’ll start by differentiating two π‘₯ to the fifth power minus nine times the sec of π‘₯ with respect to π‘₯. We can evaluate the derivative of this term by term. We can differentiate our first term by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and reduce this exponent by one. This gives us 10π‘₯ to the fourth power.

Next, to differentiate our second term, we need to recall for any real constant π‘Ž, the derivative of π‘Ž times the sec of π‘₯ with respect to π‘₯ is equal to π‘Ž times the tan of π‘₯ multiplied by the sec of π‘₯. So by setting our value of π‘Ž equal to negative nine, we can differentiate negative nine sec of π‘₯ to get negative nine tan of π‘₯ times the sec of π‘₯. This means we’ve now evaluated our innermost derivative, but we still need to evaluate our outer derivative. The first thing we need to do is rewrite our expression for our inner derivative. Doing this gives us the derivative of negative three π‘₯ cubed plus 10π‘₯ to the fourth power minus nine times the tan of π‘₯ times the sec of π‘₯ with respect to π‘₯.

And once again, we can evaluate this derivative term by term. We can differentiate the first two terms in this expression by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and then reduce this exponent by one. This gives us negative nine π‘₯ squared plus 40π‘₯ cubed. However, in our third term, we can see this is the product of two differentiable functions. So we can evaluate the derivative of this by using the product rule. We recall the product rule tells us for two differentiable functions 𝑓 and 𝑔, the derivative of 𝑓 times 𝑔 with respect to π‘₯ is equal to 𝑓 prime times 𝑔 plus 𝑔 prime times 𝑓.

We’ll set our function 𝑓 to be the tan of π‘₯, and we’ll set our function 𝑔 to be the sec of π‘₯. Now, we see to use the product rule, we need expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Let’s start with 𝑓 prime of π‘₯. That’s the derivative of the tan of π‘₯ with respect to π‘₯. And this is a standard trigonometric derivative result we should commit to memory. It’s equal to the sec squared of π‘₯. We can do the same to find 𝑔 prime of π‘₯. It’s the derivative of the sec of π‘₯ with respect to π‘₯. In fact, we’ve already discussed this. If we set our value of π‘Ž equal to one into our derivative rule, we see it’s equal to the tan of π‘₯ times the sec of π‘₯.

Now, all we need to do is substitute our expressions for 𝑓, 𝑓 prime, 𝑔, and 𝑔 prime into the product rule. And remember, we need to multiply this by negative nine. Clearing some space and then substituting these expressions into our product rule, we get negative nine times the sec squared of π‘₯ multiplied by the sec of π‘₯ plus the tan of π‘₯ times the sec of π‘₯ multiplied by the tan of π‘₯. And now we can start simplifying this expression. First we have the sec squared of π‘₯ multiplied by the sec of π‘₯. We can simplify this to give us the sec cubed of π‘₯. Similarly, we also have the tan of π‘₯ multiplied by the tan of π‘₯. We can write this as the tan squared of π‘₯. Then, all that’s left to do is distribute negative nine over our parentheses and then reorder all of our terms. And doing this gives us our final answer.

Therefore, we were able to show d by dπ‘₯ of negative three π‘₯ cubed plus d by dπ‘₯ of two π‘₯ to the fifth power minus nine times the sec of π‘₯ is equal to 40π‘₯ cubed minus nine π‘₯ squared minus nine times the tan squared of π‘₯ multiplied by the sec of π‘₯ minus nine sec cubed of π‘₯.

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