Question

derivative( (x-1)²+(y-2)²=1 )

result = http://www3.wolframalpha.com/Calculate/MSP/MSP1357519ia63i168c023i100001fdib57201d8b3hi?MSPStoreType=image/gif&s=20&w=168&h=40

wat does the y' mean?

Answer 1

y' means the derivative of y.

You should first rewrite the equation to get y on 1 side. Than you should start taking the derivative.

Answer 2

you can proceed as follows
\[2(x-1)+2(y-1)y'=0\] and then solve for
\[y'\]

Answer 3

\[(x-1)+(y-1)y'=0\]
\[(y-1)y'=-(x-1)=1-x\]
\[y'=\frac{1-x}{y-1}\]

Answer 4

so y' can also mean the gradient the tangent of the point right?

Answer 5

it is the derivative wrt x yes, so it is the formula for the slope of the line tangent to this circle.

Answer 6

oh man i had a typo everywhere sorry

Answer 7

i thought it was
\[(x-1)^2+(y-1)^2=1\] but you wrote
\[(x-1)^2+(y-2)^2=1\] so my method is write but answer is totally wrong

Answer 8

should start with
\[2(x-1)=2(y-2)y'=0\] ect