Video: Find the Derivative of a Function Defined as an Integral

Let 𝑓(π‘₯) = ∫_(2)^(π‘₯) √(2 + 3𝑑²) d𝑑. Which of the following gives 𝑓′(π‘₯)? [A] √(2 + 3π‘₯Β²) βˆ’ √(14) [B] ∫_(0.5)^(π‘₯) d𝑑/√(2 + 3𝑑²) [C] ∫_(0.5)^(π‘₯) 6𝑑/√(2 + 3𝑑²) d𝑑 [D] √(2 + 3π‘₯Β²)

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Video Transcript

Let 𝑓 of π‘₯ be equal to the integral from two to π‘₯ of the square root of two plus three 𝑑 squared with respect to 𝑑. Which of the following gives us the function 𝑓 prime of π‘₯? Option a) the square root of two plus three π‘₯ squared minus the square root of 14. Option b) the integral from 0.5 to π‘₯ of one divided by the square root of two plus three 𝑑 squared with respect to 𝑑. Option c) the integral from 0.5 to π‘₯ of six 𝑑 divided by the square root of two plus three 𝑑 squared with respect to 𝑑. Or option d) the square root of two plus three π‘₯ squared.

The question tells us that our function 𝑓 of π‘₯ is defined to be equal to the integral of some function of 𝑑 with respect to 𝑑 where the lower limit of our integral is a constant too and the upper limit of our integral is π‘₯. And the question wants us to find 𝑓 prime of π‘₯. That’s the derivative of 𝑓 of π‘₯ with respect to π‘₯. Since the question wants us to find the function 𝑓 prime of π‘₯, but 𝑓 of π‘₯ is defined as an integral, we can recall that the fundamental theorem of calculus relates differential calculus with integral calculus.

We recall that the fundamental theorem of calculus tells us that if some function lowercase 𝑔 is continuous on the closed interval from π‘Ž to 𝑏 and the function capital 𝐺 of π‘₯ is equal to the integral from π‘Ž to π‘₯ of lower case of 𝑔 with respect to 𝑑. Then one of the conclusions we can draw is that the derivative of capital 𝐺 of π‘₯ is equal to lowercase 𝑔 of π‘₯ for all π‘₯ in the open interval from π‘Ž to 𝑏. We can see that the way 𝑓 of π‘₯ is defined in the question is exactly the same as the way that capital 𝐺 of π‘₯ is defined in the fundamental theorem of calculus. It’s defined as the integral of some function with respect to 𝑑 where the lower limit is a constant and the upper limit is π‘₯.

So if we set π‘Ž to be equal to two, capital 𝐺 of π‘₯ to be equal to 𝑓 of π‘₯, and our integrand lowercase 𝑔 of 𝑑 to be equal to the square root of two plus three 𝑑 squared, then the concluding statement of the fundamental theorem of calculus in this case would state that 𝑓 prime of π‘₯ is equal to the square root of two plus three π‘₯ squared for all π‘₯ in the open interval from two to 𝑏. This would give us an expression for 𝑓 prime of π‘₯. So all we need to do is show that the first prerequisite for the fundamental theorem of calculus is true, which is that our integrand function lowercase 𝑔, which is the square root of two plus three π‘₯ squared, is continuous on some closed interval from two to 𝑏.

So to use the fundamental theorem of calculus, we need to show that the square root of two plus three π‘₯ squared is continuous on some closed interval from two to 𝑏. To do this, we first recall that if a function 𝑔 of π‘₯ is continuous and a function β„Ž of π‘₯ is continuous, then the function 𝑔 composed with β„Ž of π‘₯ is continuous on its domain. This is useful because if we set the function 𝑔 of π‘₯ to be equal to the square root of π‘₯ and the function β„Ž of π‘₯ to be equal to two plus three π‘₯ squared, then 𝑔 composed of β„Ž of π‘₯ is equal to the square root of two plus three π‘₯ squared.

So if we can show that the square root of π‘₯ is continuous and two plus three π‘₯ squared is continuous, then we would know that the function 𝑔 composed of β„Ž of π‘₯, which in this case is the square root of two plus three π‘₯ squared, is continuous on its domain. And we can actually see the domain of this function is all real numbers. Since π‘₯ squared is greater than or equal to zero, two plus three π‘₯ squared is greater than or equal to zero, so we’re just taking the square root of sum nonnegative number. We then use the following two facts about continuous functions. First, all polynomials are continuous on their domain, which is all real numbers. Second, the square root of π‘₯ is continuous on its domain.

So we’ve shown that our outer function is continuous because it’s just the square root of π‘₯. And we’ve shown that our inner function is continuous because it’s a polynomial. Therefore, we can conclude that the composition of these functions is continuous. So what we’ve shown is that our composition function the square root of two plus the π‘₯ squared is continuous on its entire domain, which we showed is actually all real numbers. So in particular, we could pick any closed interval from two to 𝑏. So we’ve shown that the first prerequisite for the fundamental theorem of calculus is also true. Therefore, we can conclude, using the fundamental theorem of calculus, that the function 𝑓 prime of π‘₯ is equal to the square root of two plus three π‘₯ squared, which was our option D.

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