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Problem 33 Easy Difficulty
Find an equation of the tangent line to the curve at the given point.
$ y = \ln (x^2 - 3x + 1), (3, 0) $
Answer
$y-0=3(x-3)$
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Top Calculus 1 / AB Educators
Kayleah T.
Harvey Mudd College
Caleb E.
Baylor University
Kristen K.
University of Michigan - Ann Arbor
Michael J.
Idaho State University
Watch More Solved Questions in Chapter 3
Video Transcript
and this problem. We're finding the equation for the tangent line to a curve. Specifically, the function of that curve is log rhythmic, and we're also given a point at which we have to find the tangent line. So the function were given is why equals the natural log of X squared minus three x plus one. So the first thing that we need to do before we can find the tangent line is find the derivative. So we need Y prime. How are we going to do that? We have to use training role. So why prime would be equal to one over X squared minus three x plus one times two X minus three. Now, if you're confused about that, basically what you do with chain rule because we know the derivative of the natural log is one over X, so we would have one over X. But in this case, it's this entire input and then we'll multiply by the derivative of that input. And then we can simplify Why? Prime to be two X minus three over X squared minus three X plus one. So we have the derivatives, but we need the slope of the tangent mind at the 0.30 so we can find why, prime of three, why crime of three would be equal to sticks minus 3/9 minus nine plus one so it would simplify to three. And now we're very close to finding the equation for a tangent line. We can use this point slope form why, minus why not equals m times X minus x dot So we can just plug in the point that were given why my zero equals three times X minus three. So you can essentially write the tangent line in this form or you can simplify it in point in slope form and we would have Why equals to reacts minus nine. So either way, they both mean the same thing. These air both the equation for the tangent line to our curve at that point. So I hope that this problem helped you understand a little bit more about how we confined the equation for a tangent line, using our knowledge of differentiation of logarithmic functions and then plugging in the point in our, um, coordinate plain to find the overall tangent line
University of Denver
Top Calculus 1 / AB Educators
Kayleah T.
Harvey Mudd College
Caleb E.
Baylor University
Kristen K.
University of Michigan - Ann Arbor
Michael J.
Idaho State University
