So let?s move on to our next example.
Having read of this question, it says, Find the equation of the tangent of the curve c
and then we have an equation.
We look at this equation and you?ll see that you see y?s and x?s mixed together.
So the first thing that we need to decide is what method we can use.
Let?s just write this down and then we?ll work through it step by step.
Our equation of c is sine y plus 4x to the power 5 equals to 10y plus 5.
It?s asking us to find the equation of the tangent to the curve at that point.
We?ll discuss what that means in a minute, but firstly, we just need to look at the nature of this equation.
Have a look that we have a y term here and a y term here.
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.
The advice here is just leave it as it is and we?ll differentiate it implicitly.
That means we?ll differentiate the entire equation in one go.
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.
So this here is my tangent. I?m trying to find the equation off this line.
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.
So if we found the gradient at 1,0, the gradient of the tangent will be constant all through.
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.
In order to find the gradient, we obviously need to differentiate.
So we?re going to dy/dx. Right, have a look at the question,
we?re going to differentiate implicitly.
I?m gonna write this down again but remember you don?t have to do this every time.
So I?m going to differentiate, what I?m trying to say here is that I?m differentiating each term separately.
So dy/dx of each term, so that we?ll remember what we?re doing.
With practiced or as we progress through, you?ll see that I?ll stop writing these
because we know that we?re differentiating, anyway.
Okay, I?m differentiating sine of y, so now I?m differentiating y with respect to x.
Remember what we?ve said. We can differentiate this as long as we multiply it with the factor dy/dx.
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.
This is okay. We are differentiating x with respect to x.
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.
So, we?re going to differentiate this with respect to y but multiply it with a factor dy/dx.
So 10y would just differentiate to 10 and then remember to multiply this with dy/dx.
Any constants by themselves would just go to zero, so that just disappears.
So this is interesting now, because we now have 2 dy/dx terms.
When I?m looking for the gradient, I?m looking for dy/dx.
So what I need to do here is to bring my dy/dx terms to one side and then work it out.
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.
You see that we now have a common factor of dy/dx so we can take that out.
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.
We can now take 10 minus cos y which to the other side and divide it.
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.
Now, we need to know the gradient at the specific point.
So remember that this is just the general gradient.
So that gradient will hold true to any part of that curve.
But we want to know the gradient at 1,0.
So you can say that at 1,0 dy/dx and you just have to substitute 1 and zero, 1 where the x is.
And zero where the y is. So I have 20 multiplied by 1 to the power of 4.
And then I have 10 minus cos of zero. That gives me 20 at the top and then cos of zero.
If you think about what cos zero is its 1, if you're not convince, just think of the cos curve.
Looks a little bit like that. And one cos is zero here, your y-value is one.
So cos of zero is one so that gives me 10 minus 1 to give me 20 over 9 and that?s your gradient.
We can now find the equation of the tangent by putting these into our equation.
So remember that the equation that we discussed of the straight line.
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.
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.
But even if you use this line it?s fine. You?ll get end up with the same answer.
So what?s our y and what?s our x? We know our y value from the curve is zero.
We know our x value from the curve is 1 and we also know our gradient is 20/9.
So it?s just a matter of putting it in. We have y minus 0 is 20/9.
X minus 1 is basically done but you can rearrange it to tidy this up.
You can move the 9 over so you?ll get 9y equals to 20x minus 20, when I multiply the 20 through.
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.