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Problem 2

In Exercises $1-20,$ find the derivative. $$y=t …

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Problem 1

In Exercises $1-20,$ find the derivative.
$$y=x \ln x$$

Answer

$y^{\prime}=1+\ln x$


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Video Transcript

So we're looking at incorporating our differentiation rules from previous chapters. And now looking at how they apply with some lager with make rules and then further into this chapter is well with, ah hyperbolic functions as well as the inverse hyperbolic functions. But to be able to do this, you need to be extremely familiar with thes differentiation rules that apply in any scenario. So you've got your differentiation. Rules involving Constance, derivative of a constant is always zero. The derivative, a constant with the function, is the constant with the derivative of the function addition. You're taking the drive of each function and adding them subtraction. You're taking the derivative of each function and subtracting them with your product rule. You've got the derivative of one multiplied by the rest, and then you're adding as you go, swapping out which function you have the derivative of of each time quotient rule very similar acceptance subtracting and we've added a squared denominator. It's the original denominator now just squared. And then, with our chain rule, we take the derivative of the outside and kind of work our way inside, taking the drivel of everything as we go and then Lastly, with our power rule, we moved the exponents the front, and go one less with our exponents. So those ones you've been working with up until now, you should have a lot of practice with them and you should be very familiar with how to apply them to a question. What's been added now are exponential in our log rhythmic functions and how they work with derivatives. So we've got the driven of R e. Not just days is e And then we have the derivative of a power with a variable exponents. We have that same power that gets left there with the derivative. But we add a natural log of the base two, round that one out. Then we also have the derivative of a natural log that just becomes one over X. If you have the natural log of a function that becomes one over your function and then following with our chain rule, we'd have to take the derivative of the function itself and multiply that on and then lastly, we have the derivative of a log of any base for that one. We take again the X, and we'd multiply it by our natural log of our base again. Even with this one, if I've got and I guess sometimes it happens not very often, but on the off chance you do have it, we could work this one out the same way. So it be won over whatever that X function is the derivative of or our natural log of a And then we'd have to multiply that again, using our chain rule to make our derivative of the inside that one doesn't happen is often. But on the off chance does we've got it. So, looking at this problem here, I've got multiplication or the product rule as well as my natural log differentiation. So I'm gonna use my product rule, which you remember. You've got your f prime g, and I'm not gonna write. My ex is in here, and then you're adding on your g prime of f. And then again, we're also using our natural log rule. So remember that um, the derivative of a natural log is just one over X. So those are the two rules were using for this one. So throwing that in here using my product rule, I'd have the derivative of X, which is just one. I'd multiply that by my second function here, and then I'd have to take my second function. So my natural log of X, I'm gonna do the druid of that which is one over X. And I'd have to multiply that by my original function, which or my f function, which is X. So cleaning this up here, I don't have to write the one out front, cause anything Times one is just one. And then my 2nd 1 here, thes exes can cancel out because one times X, divided by ex just happens to be one again. So going through that are derivative of X natural log of X would be natural log of X plus one or one plus natural log of X since with addition and action does not matter the order in which you write it

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