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Question

Solve the given problems by finding the appropriate derivative.
Find the equation of the line tangent to the curve of $y=x^{2} \ln x$ at the point (1,0).

Stanley Enemuo · Accepted Answer

So we need to find the equation of the tangent line. Why? Mhm X squared I&#x27;m like At the .1 comma zero. So to find the equation of the tangent lines that are given curves, we will first find the slope of the tangent line at the given point. So if you want the just equal to X squared X. Close two x times Alan Novak. So we have to find the slope at 1:00. Therefore putting the value of the coordinates and dy dx you get the stuff of the tangent line That is equal to one because two tom&#x27;s one Time time of one Best Equal to zero. So the equation of the line, the point slope form my mind is why why equal to m times ah minus except one to accept one Is equal to one. The wife of one is equal to zero And M is equal to one. From the above equation we get the equation of the tangent line Given the curve won&#x27;t come to zero To be 1 0 is equal to one Times X -1. So why is equal to X -1?

Clarissa Noh · Answer

Hey, it&#x27;s from someone. You rain here, so we have. Why is equal to X square minus one all over X square plus X plus one. We&#x27;re gonna use the quotient role to find the derivative when we get X square minus one The derivative X square plus X plus one minus X square minus one and the derivative of X square plus X plus one This is all over X square plus X plus one square. When we simplify this, we get X square. It was for X plus one while over at square, less X plus one square. So to find a slope of the tangent line at one comma zero we&#x27;re gonna plug in X is equal to one into our derivative when we get 2/3. So we&#x27;re gonna create a line Why is equal to 2/3 X minus one plus zero? This becomes equal to two birds. X minus 2/3

Madi Sousa · Answer

in this problem, we have to find the equation for the tangent line to a curve and specifically were given a curve that is defined by a log rhythmic function. So we have the function. Why equals X square times the natural log of X and what we need to do before we confined the tangent line is find my prime. Well, how do we do that? We&#x27;re going to have to apply the product rule. So why Prime would be equal to d X square dx times the national log of X plus x squared times d natural log of X over d x So when you do that, we simplify just a little bit. We get why prime equals two x times the natural log of X plus x So we have the derivative. But we want to know the equation for the tangent line. So what do we have to do now? We need the slope of our tangent line at the point that we&#x27;re told in the problem 10 so we can plug in our X value into y prime. So why prime would be equal to two times one times the natural log of one plus one, so we would get zero plus one, which equals one. And now we confined the tangent line very easily. We have this form. Why? Minus why not equals M times X minus X not. And then we could just plug in the X and Y coordinates from our point, so we would have y minus zero equals one times X minus one, and then we can simplify to get the tangent line equals Why equal toe X negative one x minus one. Excuse me. So I hope that this problem helped you understand how we can find the tangent line to a curb using differentiation, specifically finding the derivative of a of a logarithmic function.

Clarissa Noh · Answer

he had square. So when you read here, so we have why is equal to X to the 1/4 power minus acts. This can be rewritten as follows. So we&#x27;re gonna different she using the power roll. So we have to you. By over D X is equal to 1/4 ex to the negative 34 It&#x27;s minus one and using the slope of the tangent at 10 Do you I over D X is equal to negative three ports, and the equation of the tension is lie minus zero. It&#x27;s equal to negative 3/4 times X minus one, which is equal to negative 3/4 x flus tree forts.

Arin Asawa · Answer

So we&#x27;re giving the function f of X equals X plus X squared. Well, we&#x27;re us to do is we&#x27;re has to find the slope of the tangent line tool a curve and we&#x27;re fire as to find the equation of the tension line to the curve at negative one comma zero. And so to do that, we first have to find the slope of the tension line. Um, and so the slope of the tangent plane is going to be given by f prime of negative one. And so And we have to find out prime of negative one, using the limit definition of a derivative and so f prime of negative one to find the A derivative of a function at a point using the limit definition that&#x27;s gonna be you go to the limit as X approaches a of f of x minus f of a divided by X minus A and so f prime of negative one is gonna be cool to the limit as X approaches Negative one of fr vax minus half off negative one divided by explosive one. And so after prime of negative one is gonna be cool to the limit as X approaches. Negative one off. ATS posit squared that their function minus f of negative one. Well, that&#x27;s gonna be zero, because negative one plus one divided by X plus one. And so have prime of negative one notice if we substitute the negative one into with this expression right now, to evaluate the limit, we&#x27;re going to get a 0/0. And so we have to manipulate this expression over there, over here. And so the limit as X approaches negative one. And that&#x27;s gonna be the limit as X approaches. Negative one off, huh? Week in fact, road an X. So that&#x27;s gonna be an X Times X plus one divided by an X plus one. And so we can cancel those out. And so we&#x27;re left with f prime of negative one have prime of negative One is equal to the limit as X approaches negative one off acts which is negative one. And so that&#x27;s the slope of our attention light. So we know our attention line has formula like that. And so now we want to find RB. And to do that, we need a point on the tension line and so we know We&#x27;re finding an equation of the tension line to the curve and the point negative on comma zero. So that means the point negative one comma zero was on the 10 gin lane, and we can use it. And so negative one Commons, yours on the tension light. And so we can substitute that in and we get zero equals Ah, one plus b and so negative on equals B. And so we&#x27;re left with y equals negative eggs. Why equals negative X minus one as the slope off our

Amy Jiang · Answer

Okay, so we want to find our equation of our attention line. So let&#x27;s recall or change in line is dead equation open behind? So we&#x27;ll use our point slope form. That&#x27;s why I might is why not? Is able to m times x minus x Not so we have excellent we need to find Why not on a result. So it&#x27;s plug that into it. Tough applied at X not that&#x27;s equal to log of 10 which is equal to one. So we have that X not. Why not is equal to 10 Common one. Now let&#x27;s find our fallen derivative So F prime backs. This is equal to one over X. There&#x27;s natural log of 10. It&#x27;s plug in, or ex knots who have won over 10. The natural log 10 for a slope. So let&#x27;s put that into our equation, where why minus one is equal to 1/10. That your long of time in the next minus time. Okay, so this gives us that we have y minus one is equal to one. Actually, it&#x27;s like that as X over 10 natural law goten minus one over the natural log. 10. It&#x27;s I&#x27;d want to both sides. Okay, so we get that. Our equation of our line is why is equal to X over 10. Not your local 10 minus one over to natural 10. That&#x27;s one.

Problem

Solve the given problems by finding the appropria…

Problem 17 Easy Difficulty

Solve the given problems by finding the appropriate derivative.
Find the equation of the line tangent to the curve of $y=x^{2} \ln x$ at the point (1,0).

Answer

$$y=x-1$$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Anna Marie V.

Michael J.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$$y=x-1$$

Basic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Michael J.

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Precalculus Review - Intro

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus