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Numerade Educator

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Problem 17 Easy Difficulty

Differentiate the function.

$ T(z) = 2^z \log_2 z $

Answer

$$T^{\prime}(z)=2^{z}\left(\frac{1}{z \ln 2}+\ln z\right)$$

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Video Transcript

So for this problem, we're learning essentially how to take the derivatives of logarithmic functions. So what we're starting off in this case is a function t of Z. And it doesn't matter how we write this weaken right as TMZ ffx whatever, because just a function. Um but in this case, what we'll have is t of Z being equal to too busy times log based too, Zeke. So we're gonna want to use what's known as the product group for differentiation and how that's gonna look is we take the derivative of the first term that we're multiplying by. So to to the Z, we know that, uh, the derivative of to To the Z is going to be to To the Z I'm the natural log of to and then that will be multiplied by just log based too busy. We don't have to differentiate that at first. And then we add that with two to the scene times the derivative of this portion right here we know that the derivative of logarithmic function is going to be one over the U value. In this case, that would be easy times the natural log of the base value, which is to So that should be our answer right here. Then what we can do is simplify things further. You could either factor out the two Z. Um, that's one option or another thing that can be done is, um, simplifying things further. If we view this right here as the natural log of Z over the natural log of too, that's another way you could write it. If we do that, then this will cancel with this right here. Eso we would have that and then we can write to to the Z up top here. So this would be our final, um, mawr simplified form of it. We could also factor out of Tuesday if we wanted to. It would be too, too busy times natural log of Z plus one over z times, the natural rate of two. However, this could be an easier way of writing it. And then if we had t prime of Z, we would end up getting this exact same graph. Um but yeah, this ultimately just shows us how to dio a better job of differentiating logarithmic functions, which is crucial when dealing with applications of math which use a lot of rhythmic functions all the time.