# Logarithmic Differentiation Problem

• dkotschessaa
In summary: The slope of the tangent line at that point is 0, which means it is a horizontal line. This may seem counterintuitive, but remember that the slope of the tangent line is the instantaneous rate of change at that point. Since the function is curving upwards at that point, the slope is 0. So, in summary, the problem being discussed involves finding the equation of the tangent line to the curve y = \frac{(e^x)}{x} at the given point (1, e). The two methods being considered are differentiating first and taking logarithms first. Both methods can work but require careful use of the quotient rule and logarithm identities. Ultimately, the correct answer involves simply taking the derivative
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

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dkotschessaa said:
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.

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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.

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

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

dkotschessaa said:
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.

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.

dkotschessaa said:
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.

## What is logarithmic differentiation?

Logarithmic differentiation is a method used in calculus to find the derivatives of functions that cannot be easily differentiated using the traditional methods. It involves taking the natural logarithm of both sides of an equation and then differentiating the resulting equation.

## When is logarithmic differentiation used?

Logarithmic differentiation is used when dealing with functions that have products, quotients, or powers in them. It is also useful when dealing with functions that have variables in the exponent.

## How do you perform logarithmic differentiation?

To perform logarithmic differentiation, you first take the natural logarithm of both sides of the equation. Then, you use the logarithm rules to simplify the equation. Next, you take the derivative of both sides, using the chain rule if necessary. Finally, you solve for the desired derivative.

## What are the advantages of using logarithmic differentiation?

Logarithmic differentiation allows for the differentiation of complex functions that cannot be easily differentiated using other methods. It also simplifies the process of finding derivatives of functions with products, quotients, or powers.

## Are there any limitations to using logarithmic differentiation?

Logarithmic differentiation can be a time-consuming process and may not always result in a simplified equation. It is also not suitable for functions that do not have a defined natural logarithm.

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