Let 𝑔 be the function defined by 𝑔 of 𝑥 equals two sin of three 𝑥 plus 𝑒 to the power of cos 𝑥. Let 𝑓 be a differentiable function. The table shown gives values of 𝑓 and its derivative 𝑓 prime at selected values of 𝑥. Let 𝑦 be the function whose graph, consisting of five line segments, is shown in the given figure. Part one, find the slope of the line tangent to the graph of 𝑔 at 𝑥 equals 𝜋 by two.
In other words, we’re looking for the derivative 𝑔 prime evaluated at 𝜋 by two. To find this, we first have to find 𝑔 prime of 𝑥, which we do by using the expression we have for 𝑔 of 𝑥. We differentiate this term by term. The derivative of two sin three 𝑥 is six cos three 𝑥. We see this by using the chain rule. Setting three 𝑥 equal to 𝑢 and applying the chain rule gives us d𝑢 by d𝑥 times d by d𝑢 of two sin 𝑢. 𝑢 is three 𝑥. And so d𝑢 by d𝑥 is just three. And d by d𝑢 of two sin 𝑢 as sine differentiates to cosine is two cos 𝑢. As three times two is six and 𝑢 is three 𝑥, we get the answer six cos three 𝑥.
Now we just need to differentiate the other term, 𝑒 to the power of cos 𝑥. This is a slightly more tricky application of the chain rule. This time, we set 𝑢 to be cos 𝑥. And so we want to find d by d𝑥 of 𝑒 to the power of 𝑢. And by the chain rule, this is d𝑢 by d𝑥 times d by d𝑢 of 𝑒 to the 𝑢. What is d𝑢 by d𝑥? Well, 𝑢 is cos 𝑥. And differentiating this, we find that it is negative sin 𝑥. Cos differentiates to negative sine. Now what about d by d𝑢 of 𝑒 to the 𝑢? Well, the derivative of the exponential function is itself. So we get 𝑒 to the 𝑢. Using the fact that 𝑢 is cos 𝑥, we find the derivative in terms of 𝑥 alone. It’s negative sin 𝑥 times 𝑒 to the cos 𝑥. We therefore see that 𝑔 prime of 𝑥 is six cos three 𝑥 minus sin 𝑥 times 𝑒 to the cos 𝑥.
We want 𝑔 prime of 𝜋 by two. So we have to substitute 𝜋 by two for 𝑥. Now we can evaluate using the fact that cos of three 𝜋 by two is zero. Sin of 𝜋 by two is one. And cos of 𝜋 by two is zero. This is just negative one times 𝑒 to the zero. And as anything to the power of zero is one, this is just negative one. So the slope of the line tangent to the graph of 𝑔 at 𝑥 equals 𝜋 by two is negative one.
Part two of the question says let ℎ be the function defined by ℎ of 𝑥 equals 𝑦 of 𝑔 of 𝑥. Find ℎ prime of 𝜋 by two.
What is ℎ prime of 𝑥? Well, ℎ is this composition of functions. And how do we differentiate a composition of functions? We use the chain rule. For any two functions 𝑦 and 𝑔, the derivative of 𝑦 of 𝑔 of 𝑥 is 𝑔 prime of 𝑥 times 𝑦 of 𝑔 prime of 𝑥. We want to find ℎ prime of 𝜋 by two. So we substitute 𝜋 by two for 𝑥. Our answer involves 𝑔 prime of 𝜋 by two, which is good news. Because in part one, we found that 𝑔 prime of 𝜋 by two was negative one. ℎ prime of 𝜋 by two is therefore negative one times 𝑦 of negative one. And we can read off the value of 𝑦 of negative one from the graph of 𝑦. This graph is piecewise linear. And so it’s not too hard to see that 𝑦 of negative one is negative a third. ℎ prime of 𝜋 by two, which is negative one times this, is therefore a third. We move on to part three.
Let 𝑝 be the function defined by 𝑝 of 𝑥 equals 𝑓 of three 𝑥 times 𝑦 of 𝑥. Find 𝑝 prime of negative one.
𝑝 of 𝑥 is the product of two functions. And so to differentiate it, we have to use the product rule. The derivative of 𝑓 of 𝑥 times 𝑔 of 𝑥 is 𝑓 prime of 𝑥 times 𝑔 of 𝑥 plus 𝑓 of 𝑥 times 𝑔 prime of 𝑥. And now we have to be slightly careful because the derivative of 𝑓 of three 𝑥 is not the same thing as the derivative of 𝑓 evaluated at three 𝑥. We have to use the chain rule here. Setting 𝑢 equal to three 𝑥 and applying the chain rule, d𝑢 by d𝑥 is three. And the derivative of 𝑓 is 𝑓 prime. Recall that 𝑢 is three 𝑥. And so our derivative is three times 𝑓 prime evaluated at three 𝑥.
And to drive the point home, this is not 𝑓 prime of three 𝑥, but three times this value. So we substitute this. And the first term becomes three 𝑓 prime of three 𝑥 times 𝑦 of 𝑥. The second term is more straightforward. The derivative of 𝑦 of 𝑥 really is just 𝑦 prime of 𝑥. We don’t have to apply the chain rule.
And so now we’re ready to substitute. So we substitute and use the fact that three times negative one is negative three to simplify. Now we can read off the value of 𝑓 prime of negative three from the table. It’s negative two. And similarly, we can find the value of 𝑦 of negative one using our graph. In fact, we found it to be negative a third in the previous part of the question. We can read off 𝑓 of negative three as well from the table. It’s negative eight. And finally, what is 𝑦 prime of negative one? Well, it’s the slope of the graph of 𝑦 at negative one. And that graph is piecewise linear. We can calculate the slope using rise over run, taking the points negative three, negative one and zero, zero. The slope is the change in 𝑦 over the change in 𝑥, which simplifies to a third. So 𝑦 prime of negative one is a third. Doing the arithmetic, we find that 𝑝 prime of negative one is therefore negative two-thirds.
Part four asks us, is there a number 𝑛 in the closed interval from negative four to negative two such that 𝑓 prime of 𝑛 is negative five. Justify your answer.
Looking at our table of values, we don’t see a value of 𝑛 for which 𝑓 prime of 𝑛 is negative five. And you might think, looking at the pattern in 𝑓 prime of 𝑥, that 𝑓 prime of 𝑥 is increasing by two each time from negative four. And so will never be negative five in the interval required. However, the answer is yes. There is a number 𝑛 in that closed interval such that 𝑓 prime of 𝑛 is negative five. The mean value theorem states that if 𝑓 is differentiable on the closed interval 𝑎, 𝑏 as our function is differentiable on the closed interval negative four, negative two. Then there exists a number 𝑐 in this closed interval such that 𝑓 prime of 𝑐 is the average rate of change of the function over this closed interval. That’s 𝑓 of 𝑏 minus 𝑓 of 𝑎 all over 𝑏 minus 𝑎.
Applying this theorem to our function and the interval negative four, negative two. We see that there exists 𝑐 such that 𝑓 prime of 𝑐 is 𝑓 of negative two minus 𝑓 of negative four all over negative two minus negative four. Now we can read off the values of 𝑓 of negative two and 𝑓 of negative four. They are negative nine and one. So the numerator becomes negative nine minus one, which is negative 10. And of course, in the denominator, negative two minus negative four is just two. Negative 10 over two is negative five. And so there exists a number 𝑐 such that 𝑓 prime of 𝑐 is equal to negative five. This number 𝑐 is in the closed interval from negative four to negative two. And there’s nothing special about the letter 𝑐. We can call it 𝑛 instead. And so we have justified our positive answer to part four of this question.
Perhaps we should also write down that 𝑓 is differentiable on the closed interval from negative four to negative two, which we need for the mean value theorem to apply. We were told in the question that 𝑓 is differentiable. Period. So this is fine.