Why bother so much? Just implicitly write
and differentiate for x:
so at points where y(x) is defined,
and use to find
Ok, I have a question here, however the problem i have is more simple than the actual question itself, it's a basic math lack of understanding on my part.
Consider the curve defined by:
Find an expression for in terms of x and y, and hence give the equation of the tangent to the curve at the point (x,y) = (1,1)
So, to begin, I make:
then differentiate with respect to X:
giving
Next, differentiate with respect to Y:
giving
Now, I know that the next step is
But I don't understand where the - comes from to make the fraction negative. I know that needs to be inverted to make it into and so moving to the bottom ofthe fraction, but where does the minus sign come from?
Could someone please explain that before we move onto the next step? Thanks
I do it the other way because it is the way i learned and I can't understand the way you gave, can you help fcomplete my method? I have a lot to learn in the next two days and learning a new method from scratch isn't the best way to get that done.
I'm sorry, it's this general sort of maths notation that causes me problems. Bearwith my while i try write all you have just said out again using the full numerical functions.
I can't do general equiations/maths. Need to see the figures and values.
Ok, so when i have a fraction of a negative number over a positive one, and i take the negatiev sign out, do i invert all signs on the top?
so to
Would doing the following be the same and correct?
to
Shown the signs to explain what I mean, is this the correct manipulation?
Tell me about it, but right now i just need to able to do the papers, starting wednesday i have 3 maths papers, if i don't pass them all i fail my year, i resit the yearagain losing another year of my life, i end up being a year behind all my friends and it will cost me an extra £3000 so on the whole, i really don't want to fail these.
That being said, I don't have any more maths exams after this year.
Ok, so now I got that part, next is the next step.
"Give the equation of the tangent to the curve at the point (x,y) = (1,1)
How do i sole this last step.
The solution I am given is:
How do i find this? Thanks
Hello,
you already know the gradient of the curve at T(1, 1):
. Plug in the given values:
You are looking for the equation of a straight line passing through T (the tangent point) with a slope m = -1. Use the point slope-formula of a straight line:
If a line passes through the point with the slope m then the equation of the line is:
. Plug in all values you know (coordinates of T, slope m):
I've attached a diagram of the curve and the tangent.