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Numerade Educator

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Problem 49 Easy Difficulty

Use logarithmic differentiation to find the derivative of the function.

$ y = (\tan x)^{1/x} $

Answer

$y^{\prime}=(\tan x)^{\frac{1}{x}}\left[\frac{\sec ^{2} x}{x \tan x}-\frac{\ln (\tan x)}{x^{2}}\right]$

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Video Transcript

and this problem. We are learning how to use log rhythmic differentiation to find the derivative of a function. And you might be asking, why is log rhythmic differentiation a useful technique? Well, using log properties, we can simplify the exponents in a function very easily. So let's first review the function that were given were given. Why equals the tangent of X rays to one over X. So the first thing that we're going to do in the process of log rhythmic differentiation is will take the natural log of each side. So we'll get the natural log of y equals the natural log of the tangent of X rays to one over X And now, using natural log properties, we can essentially take this Expo in and put it as a coefficient in front of our function. So we'll get the natural log of y equals one over X times. The natural log, basically what's left the tension of X, and then what we're going to do is we'll take the derivative of each side of our equation, so we'll have one over. Why times d y d X equals negative one over x squared times The natural log of the tangent of X plus one over X times one over the tangent of X times seeking squared of X. So what we can see is this portion that I've emphasize an orange. That is where we would use the product rule because we do have a product here. But we also have to use chain rule. We have a composition of functions, and that's what I've outlined here in red. And then we can simplify this a little bit. We'll get one over Why. Times D y d X equals negative natural log of tangent of X over X squared plus one of her ex times. One over cynics, co cynics. But we can substitute this part of our equation here using a trick identity so we'll get one over y times. D y d X equals the negative natural log of tangent of X over X squared plus two over X sign two X and then we can simplify this and by multiplying, why out each side of our equation? Because, remember, we're trying to find the derivative. We don't want this one over. Why on decide with our differential, so we'll have d Y d X equals Y times two over x signed two x and minus the natural log of tangent of X over X squared. But we know why is why is the original expression the original function that we started with So we can just substitute that in and we'll find that the derivative of our function D y d X equals the tangent raised toe one over x times two over x signed two x minus the natural log of tangent of X all over X squared. So I hope that this problem help to understand not only why we use log rhythmic differentiation, but how we can go through the process of using log with Mick differentiation while we have trig and metric functions as well.

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