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Derivatives of Logarithmic Functions - Problem 2
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Recall that a tangent line is a line that touches the graph at one point. To find the equation of the tangent line at a certain point, draw the tangent line at that point. Since the slope of the tangent line is the same as the derivative of the function at that point, to find the slope of the tangent line at some x, simply plug in that value of x into the derivative (for instance, if f(x)=ln(x), recall that f'(x)=1/x; therefore, the slope of the tangent line at x=1 is f'(1)=1/1=1). Then, using the point at which you drew the tangent line, you can use point-slope form to find the equation for the line. Recall that point-slope form is y-y_{1} = m(x-x_{1}), where m is the slope you found and (x_{1},y_{1}) is the point of tangency, i.e. the point at which you drew the tangent line.
Using the tangent line of ln(x) at this point, you can approximate the natural log of other points very close to 1, such as 1.2 and 0.9, by plugging in these values for x into the equation of the tangent line. Keep in mind that this only works for values of x that are close to the point at which you drew the tangent line.

Let's try a slightly more difficult problem. We're asked to find the equation of the line tangent to y equals lnx at x equals 1, and use the tangent to approximate natural log of 1.2, and natural log of point 9.

Well, for starters let's just get the equation of the tangent line. We need two things. We need the point of tangency, and we need its slope. We have x equals 1 for the point of tangency. So all we have to do is figure out what y equals. So the y is going to be equal ln of 1. Remember that that is 0. So the point of tangency is 1, 0.

What about the slope? We need the derivative dy/dx which is 1 over x. At x equals 1, that means the slope dy/dx is 1 over 1 which is 1. So that's our slope. So we use the point-slope form; y minus equals the slope of 1 times x minus 1. That's the equation of the line in point-slope form. In slope-intercept form, it will just be y equals x minus 1.

Now that's the equation of the line. We're asked to use the tangent to approximate natural log of 1.2, and natural log of 0.9. So let's take a look at the graph of this line, and natural log and see what that means. I've got a graph drawn here. This line is our tangent line y equals x minus 1.

Now normally if we wanted to actually find the value of ln of 1.2, we plug 1.2 to natural log, and maybe use our calculator. Here, using a tangent line to approximate natural log means to plug 1.2 into this equation. This point of tangency remember is 1,0. So that's x equals 1 right there. So let's say 1.2 is here, I'll exaggerate its position a little bit. Ln of 1.2 will be the y coordinate of this point. Using the tangent line to approximate, ln of 1.2 would give me this y coordinate. They're not exactly the same, but if we stay close to x equals 1, the two values are going to be close.

So ln of 1.2 is approximately, and I use this formula, 1.2 minus 1. Let me erase this. That's 0.2. So that's an approximation for natural log of 0.2. Now just to test this approximation, let me calculate ln of 1.2 on my calculator. This also is an approximation, because, in general calculators can't give you exact values for natural log.

So natural log of 1.2, and I get 0.1823. I'll just say 0.182. That's pretty close to 0.2. Now I was also asked to approximate natural log of 0.9. So 0.9 is a little to the left of 1. So approximating this value, natural log of 0.9 is the y coordinate of this point. The approximation will be the y coordinate of this point. That's what I get when I plug 0.9 in for x here. So natural log of 0.9 is going to approximately be 0.9 minus 1 which is -0.1. This y axis is in my way. That's my approximation.

Now let's see how close it is. Natural log of 0.9 on my calculator, -0.105. So still very close. This idea of using a tangent line, to approximate a function near the point of tangency is called a linear approximation, because some functions like natural log are very difficult to compute. We sometimes use a linear approximation to get approximate values of functions. So as long as we're near the point of tangency, these approximations will be good.

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