Video: Finding the First Derivative of the Product of Two Trigonometric Functions at a Point

Given that 𝑦 = (βˆ’2 cos π‘₯ βˆ’ 7)(8 sin π‘₯ + 19), determine d𝑦/dπ‘₯ at π‘₯ = πœ‹.

03:30

Video Transcript

Given that 𝑦 is equal to negative two times the cos of π‘₯ minus seven multiplied by eight times the sin of π‘₯ plus 19, determine d𝑦 by dπ‘₯ at π‘₯ is equal to πœ‹.

The question wants us to determine d𝑦 by dπ‘₯ when π‘₯ is equal to πœ‹. And we can see that we’re given 𝑦 as the product of two functions. So we’ll differentiate this by using the product rule for differentiation, which tells us for functions 𝑒 and 𝑣, the derivative of 𝑒 times 𝑣 with respect to π‘₯ is equal to 𝑣 times d𝑒 by dπ‘₯ plus 𝑒 times d𝑣 by dπ‘₯. So to apply the product rule to 𝑦, we’ll set 𝑒 equal to negative two times the cos of π‘₯ minus seven and 𝑣 equal to eight times the sin of π‘₯ plus 19.

Now, to apply the product rule, we’re going to need to find expressions for d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯. We’ll start with d𝑒 by dπ‘₯. That’s the derivative of negative two times the cos of π‘₯ minus seven with respect to π‘₯. We’ll differentiate this term by term. We recall for any constant π‘Ž, the derivative of π‘Ž times the cos of π‘₯ with respect to π‘₯ is equal to negative π‘Ž times the sin of π‘₯. This gives us the derivative of negative two times the cos of π‘₯ is equal to negative one times negative two times the sin of π‘₯. And we know the derivative of the constant negative seven is equal to zero.

We can simplify negative one times negative two times the sin of π‘₯ to just be two sin π‘₯. We want to do something similar to find d𝑣 by dπ‘₯. That’s the derivative of eight times the sin of π‘₯ plus 19 with respect to π‘₯. Again, we want to differentiate this term by term. We recall for any constant π‘Ž, the derivative of π‘Ž times the sin of π‘₯ with respect to π‘₯ is equal to π‘Ž times the cos of π‘₯.

So we can differentiate eight times the sin of π‘₯ with respect to π‘₯ to get eight times the cos of π‘₯. And 19 is a constant, so its derivative is equal to zero. So we’re now ready to find an expression for d𝑦 by dπ‘₯ by using the product rule. We have that d𝑦 by dπ‘₯ is equal to 𝑣 times d𝑒 by dπ‘₯ plus 𝑒 times d𝑣 by dπ‘₯. Substituting in our expressions for 𝑒, 𝑣, d𝑒 by dπ‘₯, and d𝑣 by dπ‘₯, we get that d𝑦 by dπ‘₯ is equal to eight sin π‘₯ plus 19 multiplied by two sin π‘₯ plus negative two times the cos of π‘₯ minus seven multiplied by eight cos of π‘₯.

And we might be tempted to start simplifying this expression. However, remember the question only wants us to find d𝑦 by dπ‘₯ when π‘₯ is equal to πœ‹. So we can just substitute π‘₯ is equal to πœ‹ directly into this expression. Substituting π‘₯ is equal to πœ‹ into this expression, we get that d𝑦 by dπ‘₯ at πœ‹ is equal to eight sin πœ‹ plus 19 multiplied by two sin πœ‹ plus negative two cos of πœ‹ minus seven times eight cos of πœ‹.

And we can evaluate this expression. We know that the sin of πœ‹ is equal to zero and the cos of πœ‹ is equal to negative one. So in our first term, we’re multiplying by zero. So our entire first term evaluates to give us zero. In our second term, we have negative two multiplied by negative one is equal to two. Then we subtract seven for this and multiply this by eight times negative one, which is negative eight. And we can calculate this to give us negative five multiplied by negative eight, which is equal to 40.

Therefore, we’ve shown if 𝑦 is equal to negative two times the cos of π‘₯ minus seven multiplied by eight times the sin of π‘₯ plus 19, then d𝑦 by dπ‘₯ at π‘₯ is equal to πœ‹ is equal to 40.

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