# Differentiation of Exponentials and Logs: Exercise 4

by Batool Akmal

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DLM Differentiation of Exponential and Logarithmic Functions Exercise Calculus Akmal.pdf
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00:01 And now the final question, we're looking at y equals to ln of 3x plus 1 and it's been divided by e to the 2x. Lots of thinking to do for this question.

00:11 So let's just write it out, we have y equals to ln of 3x plus 1, over e to the power of 2x.

00:20 Now, you look at it and you recognize what kind of function that is, you can see that you have a function at the top, and then you also have a function at the bottom, so instantly we're thinking that we've to use the quotient rule.

00:36 Now, you have two options, you can either use the quotient rule, or like I mentioned before a lot of times you can actually rewrite this if I take this up as e to the minus 2x multiplied by ln of 3x plus 1.

00:50 You can also use the product rule, it really is up to you on the function that this is possible on, it's really a matter of what you prefer and what you find easier.

01:00 So just so we practice some quotient rule, and we'll use the quotient rule here, but it might be easier just using this form of it and use the product rule.

01:11 So the quotient rules states that dy/dx is vdudx minus udvdx over v squared.

01:19 Let's try this out here, if we split this into u is ln of 3x plus 1 and v is e to the 2x, if we differentiate ln we got 1/3x plus 1 and don't forget to multiply it with the differential of the inside which is 3.

01:40 When we differentiate v dash, remember that it's the function of a function so e to the 2x stays as it is, but then you also multiply it with 2.

01:52 Let's put it together into the formula, so we have vdudx minus udvdx so we'll have v multiplied by the dy/dx, so that gives me e to the 2x, multiplied by 3 over 3x plus 1.

02:12 You then minus it, with these two terms here, so we're now doing udvdx, so we have 2e to the 2x and this is being multiplied with ln of 3x plus 1.

02:28 So now we divided it by v squared so that's e to the 2x all to the power of 2.

02:35 You can see that we've done the differentiation, we've kind of done the difficult part, now it just the matter of simplifying it and using a little bit of algebra, if I come up here just to make some more space, you can multiply these two values at the top so you end up with 3e to the 2x all over 3x plus 1 and you then have minus 2e to the 2x multiplied with ln of 3x, plus 1, and it's all over e to the 4x. And remember when you have to power raise by another power, you multiply the powers so it's e to the power of 4x.

03:17 Okay, we can take 3x plus 1 as a common denominators, so we can put this over 3x plus 1 as well, but in order to do that you'd have to multiply the top with 3x plus one as well.

03:31 So 3x plus 1 and the bottom with 3x plus 1, you can treat that as fraction, as something being over 1.

03:38 So let's just do that, we can take 3x plus 1 as a common denominator, and that's just for this term here, the nominator, so we take 3x plus 1 as a common denominator, we have 3e to the 2x, minus, I have 2e to the 2x, multiplied by 3x plus one, multiply by ln of 3x plus 1. And this is also all over e to the 4x.

04:09 Now, the easiest thing to do is to combine these two denominators because they're together at the bottom, there's not much more they can simplify, you could take a factor of e to the 2x out, if that helps, from this term here, so we could take an e to the 2x out just to simplify, so we could take e to the 2x out leaving us with 3 minus 2, 3x plus 1 and ln of 3x plus 1 and this is all over e to the 4x multiplied by 3x plus 1.

04:48 There's a final little thing that you could do here and that is simplify your e to the 2x this term with e to the 4x so this term here with this term there, just because they're common in both terms so if you simplify this you can cancel e to the 2x and that here will become 2x, so if I just change that here just to change, leave some more space, so that e to the 4x now becomes e to the power of 2x, and that's your final answer. So 3 minus 2 brackets 3x plus 1, ln of 3x plus 1, and it's all being divided by e to the 2x brackets 3x plus one.

The lecture Differentiation of Exponentials and Logs: Exercise 4 by Batool Akmal is from the course Differentiation of Exponential and Logarithmic Functions.

### Included Quiz Questions

1. dy/dx = [1 - ln(x)] / x²
2. dy/dx = [1 + ln(x)] / x²
3. dy/dx = ln(x) / x²
4. dy/dx = [ln(x) - 1] / x²
5. dy/dx = [1 - ln(x)] / x
1. dy/dx = [1 - xln(x)] / [xeˣ]
2. dy/dx = [1 + xln(x)] / [xeˣ]
3. dy/dx = [1 - xln(x)] / eˣ
4. dy/dx = [1 - ln(x)] / (xeˣ)
5. dy/dx = [1 - ln(x)] / eˣ
1. [eˣ - (eˣ + 1)ln(eˣ + 1) ] / [eˣ (eˣ + 1)]
2. [1 + (eˣ + 1)ln(eˣ + 1) ] / [eˣ (eˣ + 1)]
3. [1 - ln(eˣ + 1) ] / [eˣ (eˣ + 1)]
4. [eˣ - (eˣ + 1)ln(eˣ + 1) ] / (eˣ + 1)
5. [1 - ln(eˣ + 1) ] / eˣ

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