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# Differentiate the function.$y = \log_2 (x \log_5 x)$

## $\frac{\ln x+1}{\ln 2(x \ln )}$

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Differentiation

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### Video Transcript

So for this problem right here, what we're given is a log, rhythmic expression or log rhythmic equation function. Rather, that's why equals large base to of X log. Based five of acts on DWhite we often find in math is that logs are tricky toe to deal with. We don't like dealing with vlogs. Um, they're messy. They have all these different bases, so we don't want to deal with logs instead, What we want to deal with is natural logs. They're a lot nicer to deal with. So that's what we're gonna do. We're going to convert this into a natural log function. So what we have is y equals the, um, natural log of ex Natural log X over the natural log of five. This is all over the natural log of to And the reason we know this to be the case is because log base A of B is equal to the natural log of be over the natural log of a. So we have log base tomb down here. So that's why we get natural log of two down there. We have the natural log of B up here because that is the B value right there. So with all that settled, we can now simplify things further using our log properties. So with the log properties, we can write this out now as why equals the natural log of X natural log X minus natural log five all over natural log of to then this is the same thing as one. Over the natural log of to times the natural log of X natural log X minus log based too of five. So we're going back. Thio log form Partially. So now when we differentiate with respect to X Um, what we can dio is the chain role. This is why equals So we'll get why Prime being equal to one over natural log of two. That's just a constant times the derivative of this portion. Right here we'll end up giving us one over ex natural Log X, But then we have to do the chain real procedure. Uh, so with chain role, we're going to have our own little product role inside, which is going to be natural log of X plus one over X Times X. That's just gonna be one. So the way that we can rewrite this is natural log of X plus one and then we're going to have minus zero because this is just a constant right here. So that's just gonna be zero when we differentiate. So our final answer that will end up getting now is going to be why prime equals one plus the natural log of X over X natural log of to natural log of X, and this will be our final answer for the problem.

California Baptist University

Derivatives

Differentiation

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