Video: AP Calculus AB Exam 1 • Section I • Part A • Question 14 | Nagwa Video: AP Calculus AB Exam 1 • Section I • Part A • Question 14 | Nagwa

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Video: AP Calculus AB Exam 1 • Section I • Part A • Question 14

Find 𝑓′(𝑥) if 𝑓(𝑥) = cot 𝑥 − csc² 𝑥.

02:33

Video Transcript

Find 𝑓 prime of 𝑥 if 𝑓 of 𝑥 is equal to cot 𝑥 minus csc squared 𝑥.

Here, we have a trigonometric function in 𝑥 for which we’re looking to find the derivative. That’s 𝑓 prime of 𝑥. We can differentiate each part of the expression separately. To find the derivative of cot of 𝑥 or the cotangent of 𝑥 with respect to 𝑥, we use the standard result. That says that the derivative of cot of 𝑥 with respect to 𝑥 is negative csc squared 𝑥.

But what about the derivative of csc squared of 𝑥? We’re going to rewrite csc squared of 𝑥 as csc 𝑥 squared. Remember these mean exactly the same thing. It’s just notation. But by writing like this, we can see that we have a function of a function. It’s a composite function. And this means we can use the chain rule to differentiate csc 𝑥 squared.

The chain rule says that if 𝑦 is a function of 𝑢 and 𝑢 is a function of 𝑥, then the derivative of 𝑦 with respect to 𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. So we let 𝑢 be equal to csc 𝑥. And that means 𝑦 is equal to 𝑢 squared. To use the chain rule, we’re going to need to differentiate 𝑢 with respect to 𝑥 and differentiate 𝑦 with respect to 𝑢.

Once again, we recall a standard result. The derivative of csc 𝑥 with respect to 𝑥 is negative csc 𝑥 cot 𝑥. So our d𝑢 by d𝑥 is negative csc 𝑥 cot 𝑥. The derivative of 𝑢 squared with respect to 𝑢 is two times 𝑢 to the power of one or just two 𝑢. This means then that the derivative of csc squared 𝑥 with respect to 𝑥 is negative csc 𝑥 cot 𝑥 times two 𝑢.

But we stated that 𝑢 was equal to csc 𝑥. So we replace 𝑢 with csc 𝑥. And the derivative of csc squared 𝑥 with respect to 𝑥 is negative two csc squared 𝑥 cot 𝑥. So 𝑓 prime of 𝑥 is negative csc squared 𝑥 minus negative two csc squared 𝑥 cot 𝑥. Well, a negative multiplied by a negative is a positive. So we rewrite this a little bit. And then, we spot that we have a common factor of csc squared 𝑥.

We fully factor our expression and we see that 𝑓 prime of 𝑥 the derivative of our function with respect to 𝑥 is csc squared 𝑥 multiplied by two cot 𝑥 minus one.

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