Problem

(a) If $ f(x) = x \sqrt {2 - x^2}, $ find $ f'(x)…

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Problem 56 Hard Difficulty

(a) The curve $ y = \mid x \mid /\sqrt {2 - x^2} $ is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point (1, 1).
(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

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Video Transcript

Okay, here we have the function. Why equals the absolute value of X over the square root of two minus X squared. And we want to find the equation of the tangent line at the 0.11 Now, keep in mind that at that point, the absolute value of X is positive and the absolute value of X is equal to X when you have a positive X value. So we can just replace the absolute value of X with X when we're at a positive X value. And that makes the differentiation much simpler. So now, to find the derivative, we're going to use the quotient rule. So what we see here is the bottom times, the derivative of the top, minus the top times, the derivative of the bottom. And here we're using the chain rule. So because we have a square root function, it's a 1/2 power function. So we bring down the 1/2 and we raise the inside to the negative 1/2 and then we multiply by the derivative of the inside, and then it's over, the bottom squared. So now we can go ahead and simplify that a little bit changing those 1/2 powers back to square roots and the negative 1/2 power part went down to be a denominator right there. Okay, we're finding the derivative at the 0.11 That's going to be the slope of the tangent line. So let's go ahead and substitute one in for X and things simplify quite a bit. We see here we have things like the square root of two minus one squared. So that's going to be the square root of one, which is one we have that again down here, another square root of one, which is one and even in the denominator, we just have one. So we have a whole bunch of ones. So the derivative simplifies to be to the derivative at one is to remember that that is the slope of the tangent line. So now we can use point slope form with our 0.11 and our slope to substitute those into point slope form and we get why minus one equals two times the quantity X minus one we can distribute the to, and then we can add one to both sides and we have y equals two x minus one as our tangent line equation. So next what we want to do is graph both the function and the tangent line on the calculator. So we go to Why equals we type in the function in the UAE one line and we type inthe e tangent line is why, too. And then we're going to graph thes and we'll see that that truly is the tangent line at that point. So for a window, I chose to use X values from negative to to to and why values from negative 2 to 5. And here's the graph. What we see in blue is the bullet nose function. Now we see why it's called that, and what we see in red is the tangent line, and we see the point of tangent. C is indeed X equals one. That's the 0.11

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Calculus: Early Transcendentals

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Problem

(a) If $ f(x) = x \sqrt {2 - x^2}, $ find $ f'(x)…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals