# Parametric Differentiation: Exercise 3

by Batool Akmal
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00:01 Let’s have a look at our next example. We’re building up in difficulty, so let’s try this out.

00:06 We’ve got two functions. We can see that y equals to te to the minus 3t and x = 3t² - 1.

00:18 Firstly, look at your equations. You identify that it is y in terms of t’s and then you’ve got x in terms of t. So, we have a parametric set of equations. We now need to differentiate them.

00:32 Now, there’s another thing that you need to observe here. When you look at your y function, you hopefully will notice that you have t and then you have another function of t.

00:42 So basically, you have two functions of t that are timesing together. So, you have t that’s multiplying with e to the -3t. This isn't one function. Quite a common mistake because it looks like quite a concise little function but they are two functions timesing together.

00:58 So, we’ll have to be a little bit careful when we differentiate this. Dy/dt, we’re using the product rule.

01:05 Now over time, I’m hoping that you are all using the product rule a little bit faster.

01:11 So rather than writing it all out, you're just using the fact that you leave one function as it is, differentiate the other and then you do the same to the other function. But I’ll just remind you what that is. Dy/dx using the product rule is udvdx + vdudx. So, rather than writing the whole rule out, I’m just going to try and do this here. The product rule, if I leave t as it is and then now, I differentiate e to the -3t. So if I have a function y equals to e to the -3t, the differential of this, look at it carefully. It’s a function and another function. So, it’s a function inside of a function. So when I differentiate this, I should write it with respect to t, not with respect to x.

01:59 When you differentiate this, you are going to leave e as it is because e to the power of anything differentiates to the same. It doesn't change. So, e to the -3t stays as e to the -3t.

02:13 Then you multiply it with the differential of -3t which is just -3. So, when I differentiate e to the -3t, my answer is -3e to the -3t. Then remember, I still need to do the other way around. So now, I can differentiate t which just gives me 1. Then e to the -3t can stay as it is. Let’s just tidy this up. That gives me -3te to the -3t + e to the -3t.

02:46 We’ll see what we need to do but we can just factorize it here by taking a common factor of e to the -3t out writing it as -3t +1. We’ll see if that was necessary shortly when we put it together. Meanwhile, let’s come back to dx /dt, so we still need to do this.

03:07 Dx/dt is a bit more straightforward than the other function. Bring the power down. Decrease the power by 1.

03:14 So, that just gives you 6t and any constant just disappears. Remember for our definition, we’re going to need dt/dx. So, that gives me 1 over 6t. Let’s put it all in to our definitions.

03:31 So dy/dx states you have dy/dt multiplied by dt/dx. So dy/dt is this function here.

03:44 So, we’ve got e to the -3t multiplied by -3t + 1. Then we’re multiplying it all with 1 over 6t.

03:53 Unfortunately, there’s not much you can cancel out here. So we can rewrite this as e to the -3t -3t + 1 all over 6t. That is your gradient, dy /dx.

The lecture Parametric Differentiation: Exercise 3 by Batool Akmal is from the course Parametric Differentiation.

### Included Quiz Questions

1. -2t(e^t)/5
2. 2t(e^t)/5
3. t(e^t)/5
4. -t(e^t)/5
5. (e^t)/5
1. 2
2. 1
3. -2
4. -1
5. -e^t
1. [t+1] / [e^(-t) (2t+1)]
2. [t+1] / [e^(t)(2t+1)]
3. [t+1] / [e^(t)(2t-1)]
4. [t] / [e^(t) (2t+1)]
5. [t-1] / [e^(-t)(1-2t)]

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