# Implicit Differentiation: Example 2

by Batool Akmal
(1)

My Notes
• Required.
Learning Material 2
• PDF
DLM Implicit Differentiation Calculus Akmal.pdf
• PDF
Report mistake
Transcript

00:01 So let?s move on to our next example.

00:04 Having read of this question, it says, Find the equation of the tangent of the curve c and then we have an equation.

00:11 We look at this equation and you?ll see that you see y?s and x?s mixed together.

00:16 So the first thing that we need to decide is what method we can use.

00:19 Let?s just write this down and then we?ll work through it step by step.

00:23 Our equation of c is sine y plus 4x to the power 5 equals to 10y plus 5.

00:36 It?s asking us to find the equation of the tangent to the curve at that point.

00:41 We?ll discuss what that means in a minute, but firstly, we just need to look at the nature of this equation.

00:47 Have a look that we have a y term here and a y term here.

00:50 You might be tempted to rearrange it but it would be really difficult to do this because you?ve got sine of y and 10y.

00:58 The advice here is just leave it as it is and we?ll differentiate it implicitly.

01:04 That means we?ll differentiate the entire equation in one go.

01:07 Secondly, let?s come back to the question. So this is some type of curve, I don?t know what it looks like. We?ll just say, ?That?s our curve.? It says, ?Find the equation of the tangent to the curve c at the point 1,0.? So imagine this is my point here, 1,0 and it?s asking me to find the equation of the tangent.

01:27 So this here is my tangent. I?m trying to find the equation off this line.

01:33 Remember what we said previously that the equation of the tangent next to the curve will share the gradient of that curve at that specific point.

01:42 So if we found the gradient at 1,0, the gradient of the tangent will be constant all through.

01:48 It would be just the same number. So we just have to find the gradient here, at this point and then we can use that for the tangent.

01:55 In order to find the gradient, we obviously need to differentiate.

01:59 So we?re going to dy/dx. Right, have a look at the question, we?re going to differentiate implicitly.

02:06 I?m gonna write this down again but remember you don?t have to do this every time.

02:10 So I?m going to differentiate, what I?m trying to say here is that I?m differentiating each term separately.

02:16 So dy/dx of each term, so that we?ll remember what we?re doing.

02:26 With practiced or as we progress through, you?ll see that I?ll stop writing these because we know that we?re differentiating, anyway.

02:33 Okay, I?m differentiating sine of y, so now I?m differentiating y with respect to x.

02:39 Remember what we?ve said. We can differentiate this as long as we multiply it with the factor dy/dx.

02:45 So look back at your trig, sine y differentiates to cos y but because I differentiated y, I must multiply it with the factor dy/dx. That?s done. Let?s move on to our next term.

02:59 This is okay. We are differentiating x with respect to x.

03:03 So just differentiate that as you would. So that gives me 20x to the 4, equals next term, observe we?ve got a y term again.

03:12 So, we?re going to differentiate this with respect to y but multiply it with a factor dy/dx.

03:18 So 10y would just differentiate to 10 and then remember to multiply this with dy/dx.

03:24 Any constants by themselves would just go to zero, so that just disappears.

03:30 So this is interesting now, because we now have 2 dy/dx terms.

03:35 When I?m looking for the gradient, I?m looking for dy/dx.

03:39 So what I need to do here is to bring my dy/dx terms to one side and then work it out.

03:47 You can take it to any side that you want. If you want to leave this positive, if we do 20x to the 4 equals to 10 dy/dx and then you minus cos y, dy/dx.

04:01 You see that we now have a common factor of dy/dx so we can take that out.

04:06 So we can write it as 20x to the 4 equals to dy/dx is your common factor, leaving you with 10 minus cos of y on the inside because 10 minus cos of y is multiplying.

04:19 We can now take 10 minus cos y which to the other side and divide it.

04:24 So we have 20x to the 4 and then we are dividing it with 10 minus cos y to give us our dy/dx and so we?ve found the gradient of this rather complicated-looking curve.

04:42 Now, we need to know the gradient at the specific point.

04:44 So remember that this is just the general gradient.

04:47 So that gradient will hold true to any part of that curve.

04:50 But we want to know the gradient at 1,0.

04:53 So you can say that at 1,0 dy/dx and you just have to substitute 1 and zero, 1 where the x is.

05:03 And zero where the y is. So I have 20 multiplied by 1 to the power of 4.

05:10 And then I have 10 minus cos of zero. That gives me 20 at the top and then cos of zero.

05:19 If you think about what cos zero is its 1, if you're not convince, just think of the cos curve.

05:26 Looks a little bit like that. And one cos is zero here, your y-value is one.

05:32 So cos of zero is one so that gives me 10 minus 1 to give me 20 over 9 and that?s your gradient.

05:44 We can now find the equation of the tangent by putting these into our equation.

05:50 So remember that the equation that we discussed of the straight line.

05:53 So if we just talked about this is y equals to mx plus c or the method that we were using was the faster method, so we were using this equation, y minus y1 equals to mx minus x1.

06:06 So that is the same equation but just a little bit simpler, you don?t have to find your c, substitute values and then find your c.

06:14 But even if you use this line it?s fine. You?ll get end up with the same answer.

06:19 So what?s our y and what?s our x? We know our y value from the curve is zero.

06:25 We know our x value from the curve is 1 and we also know our gradient is 20/9.

06:31 So it?s just a matter of putting it in. We have y minus 0 is 20/9.

06:38 X minus 1 is basically done but you can rearrange it to tidy this up.

06:44 You can move the 9 over so you?ll get 9y equals to 20x minus 20, when I multiply the 20 through.

06:53 So that is essentially your y equals to mx plus c. If you want to get y by itself, you can do y equals to 20/9x minus 20/9 so I divided the other side by 9 and now you can see, if you compare it with this equation y equals mx plus c that your gradient is 20/9 because that?s your m and your intercept is minus 20/9.

The lecture Implicit Differentiation: Example 2 by Batool Akmal is from the course Implicit Differentiation.

### Included Quiz Questions

1. dy/dx=[55-cos(x)] / 2
2. dy/dx=[55+cos(x)] / 2
3. dy/dx=55-cos(x)
4. dy/dx=[55x-cos(x)] / 2
5. dy/dx=[55x+cos(x)] / 2
1. 28
2. 56
3. 55/2
4. 55
5. 0
1. 1
2. 0
3. -1
4. 2
5. -2
1. y = (7/6) x + (13/6)
2. y = (1/6) x + (13/6) .
3. y = (7/6) x - (13/6) .
4. y = (1/6) x - (13/6) .
5. y = (7/3) x + (13/3)

### Customer reviews

(1)
5,0 of 5 stars
 5 Stars 5 4 Stars 0 3 Stars 0 2 Stars 0 1  Star 0