Logarithmic Differentiation Problem

[dkotschessaa]

The problem I am having with these is that I'm sometimes not sure whether to begin differentiating first, or to start taking logarithms - and whether the order changes the outcome

For example:

Find the equation of the tangent line to the curve at the given point.

$$y= \frac{(e^x)}{x}$$

At the points (1, e)

If I differentiate first I end up with $$y' = \frac{(e^x - 1)}{y}$$. Plugging in the values to get the slope, I get $$m = \frac{(e-1)}{e}$$

If I take logarithms of both sides prior to differentiating I get

$$dx/dy = y[\frac{1}{e^x} - 1(x)]$$

which would give me a slope $$\frac{(e^2 - e)}{e}$$

All of which are rather ugly things to be plugging into the point-slope form of the tangent line equation.

I suspect I'm doing something in the wrong order or missing a larger point.

-Dave K

 

[spamiam]

Either method will work. Unfortunately, both of your answers are incorrect.

If I differentiate first I end up with $$y' = \frac{(e^x - 1)}{y}$$.

If you differentiate first, make sure that you use the quotient rule correctly. How did you get a y on the right-hand side??

If you take the logarithm first, you should get

$$\ln y = \ln\left(\frac{e^x}{x}\right).$$

From there you can use some logarithm identities (e.g., $\ln(a/b) = \ln a - \ln b$) and implicit differentiation to get the answer.

 

[gb7nash]

You don't need logarithms (I'm not sure why you're using them). As you have stated in the problem, we have an explicit formula for y. All we need to do is take the derivative of y with respect to x (as spamiam has stated, you need to make use of the quotient rule). Plug in x = 1 to find your slope.

After that, make use of y = mx + b and solve for b.

 

[dkotschessaa]

Ok, maybe I am making things overly complicated. When I used the quotient rule only (That was another attempt) I end up with y'(1) = 0. If it had a slope of 0 how can it be curving upwards? (I know the natural log behaves strangely sometimes, so perhaps this is correct it's certainly not intuitive) The the equation of the tangent line comes out to $$y - e = 0(x-1)$$ or $$y = e$$

-DaveK

 

[dkotschessaa]

Ok, was looking at the wrong place on my graph. It makes sense that the slope at this point is 0, and that the equation would then be y = e

 

[gb7nash]

Ok, was looking at the wrong place on my graph. It makes sense that the slope at this point is 0, and that the equation would then be y = e

Correct. You got it.

 

[dkotschessaa]

Thanks for the help all. This is a 6 week calc I summer course. By the time we're through a chapter I am just beginning to grok the previous.

 

[Vonalak]

Ok, maybe I am making things overly complicated. When I used the quotient rule only (That was another attempt) I end up with y'(1) = 0. If it had a slope of 0 how can it be curving upwards? (I know the natural log behaves strangely sometimes, so perhaps this is correct it's certainly not intuitive) The the equation of the tangent line comes out to $$y - e = 0(x-1)$$ or $$y = e$$

-DaveK

Yes, that would indeed be correct.