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Problem 18 Easy Difficulty

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

$ \displaystyle y = \int^1_{\sin x} \sqrt{1 + t^2} \,dt $

Answer

$$
-\cos x \sqrt{1+\sin ^{2} x}
$$

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Video Transcript

yeah. Were given a function and were asked to use part one of the fundamental theorem of calculus to find the derivative of this function, the function is Y equals the integral from the sine of X. Yes, 21 of square root of one plus T squared. Mhm. GT Mhm. Yeah, recall what part one of the fundamental theorem of calculus says this is that if a function little F is continuous on a closed interval, how old and the function big F. Which is the integral from one end of that closed interval two X. Of little F. Then the derivative of big F F. Prime of X is little F of X. So, first of all thanks. It is true that the square root of one plus T square is continuous for all values of T. We can also Use integral properties to rewrite this as the opposite of the integral from 1 to the sine of X. Of the square root of one plus T squared D. T. And now using part one of the fundamental theorem, it follows that G. Prime not prime that I guess I would say why prime of X. Black? Well this is the opposite of the derivative of the integral which using the chain rule this is going to be yeah um the derivative of the integral from 12 X. Of the square one plus T squared, which is simply the square root of one plus X squared times the derivative of the inside sign index, which is co sign of X. Which we could also rewrite as get a couple of cents for this. Right? I'm sorry. Instead of one plus X squared, we substitute Synnex for X here. So this is squared of one plus sine squared of X. And so this is negative co sign of X times the square root of one plus sine squared of X. That's what more people do.