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Find the derivative of the function. Simplify where possible.$ y = x \sin^{-1} x + \sqrt{1 -x^2} $

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$$\sin ^{-1} x$$

00:49

Frank Lin

00:58

Doruk Isik

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 5

Implicit Differentiation

Derivatives

Differentiation

Leo L.

February 19, 2022

As noted before, Madi S is a great teacher. Clear step by step expression and neat writing !

Missouri State University

Harvey Mudd College

University of Nottingham

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

00:44

Find the derivative of the…

02:12

03:19

Find the derivative of $y$…

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0:00

01:04

in this problem. We are practicing our differentiation skills specifically with using inverse triggered a metric functions. And also, this is a great example of in very informal proof. And this is a great precursor to future math. Classes that you might take were given the function y equals x times, the inverse sign of X plus the square root of one minus X squared. So we have to differentiate this to prove that we get the inverse sign of X. We're told that's what we have to wind up with. So why prime are derivative would be equal to the derivative with respect to X of X sine inverse X plus the derivative of the square root of one minus X squared. So how am I going to do this? Well, this first term looks like a product, so we're going to use the product rule and the second term we're going to use chain rule. So when we apply, those rules will get that Why prime equals the inverse sign of X times one plus x times one over the square root of one minus X squared, plus 1/2 times the square root of one minus X squared times negative two x So hopefully that made sense how we went from product ruled chain rule to get this derivative here. And then we could simplify it. We could get why prime equals the inverse sign of X plus X over the square root of one minus X squared minus X over the square root of one minus X squared. While these two terms are the exact staying with opposite signs so they can cancel. So we're going to get that. Why Prime equals the inverse sign of X, just as we were supposed to. The problem tells us that that's what we have to find. So I hope that this problem helped you understand a little bit more about differentiation, how we can choose the rules to differentiate. And then I hope this helped you understand a little bit more about about an informal proof, or maybe what to do when a problem tells you to prove something

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