🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Oh no! Our educators are currently working hard solving this question. In the meantime, our AI Tutor recommends this similar expert step-by-step video covering the same topics. Numerade Educator ### Problem 2 Easy Difficulty # Differentiate each of the following:a.$y=e^{3 x}$b.$s=e^{3 t-5}$c.$y=2 e^{10 t}$d.$y=e^{-3 x}$e.$y=e^{5-6 x+x^{2}}$f.$y=e^{\sqrt{x}}$### Answer ## a.$3 e^{3 x}$b.$3 e^{3 t-5}$c.$20 e^{10 t}$d.$-3 e^{-3 x}$e.$(-6+2 x) e^{5-6 x+x^{2}}$f.$\frac{1}{2 \sqrt{x}} e^{\sqrt{x}}\$

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{'transcript': "problem number 27 Section 45 Again, Everything in this section deals with finding derivatives of functions that involve logarithms. So here I have. Why is equal to e to the five X divided by the natural log of three x each of the five x over the natural log of three to the X. So again, guiding principle here I've got something that looks like y is equal to F over G. Why prime is quotient rule F prime G minus G prime f over cheese squared. In my particular case, eat of the five x represents F and natural law. Go three x represents a g. So to go find this derivative. So why Prime is going to be equal to F Prime G. So the derivative eat of the five X is five e to the five X so that's F primes, Then times G. That's gonna be times natural Log of three x minus G prime f So g prime f. That is gonna be what one over three x times 32 That's G prime. All of that times F, which is either the five x over G squared, which is Ellen of three x squared and it looks like I can simplify a bit by multiplying everything by five X. And so I multiply everything by five acts she meet motive, everything by X because what you're going to see here? I've got three over three I can keep from having a complex fraction if I multiply everything so by X so I'm gonna have five x e to the five x natural log of three X and then minus E to the five x over X natural log of three X and then all of that is squared and it looks like I could probably factor out and eat to the five X, um, for a further simplification. So if I were to do that, I'm gonna have e to the five X and then I'm left with five x natural log of three X minus one, all of that divided by X natural log of three x squared. So each of the five x five x of natural log of three X minus one over X natural log of three x All that quantity squared"}

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