Thread: find the slope of the curve

The equation sin xy = y defines y implicitly as a function of x.
Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

i know that y' = (ycos(xy))/(1-xcos(xy))

but i just dont know how to solve the problem when i plug in x and and y to find the slope

please help

ty

Originally Posted by vtong

The equation sin xy = y defines y implicitly as a function of x.
Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

i know that y' = (ycos(xy))/(1-xcos(xy))

but i just dont know how to solve the problem when i plug in x and and y to find the slope

please help

ty

y=sin(xy) so and plug in to get which is the slope of your function

Originally Posted by vtong

The equation sin xy = y defines y implicitly as a function of x.
Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

i know that y' = (ycos(xy))/(1-xcos(xy))

but i just dont know how to solve the problem when i plug in x and and y to find the slope

please help

ty

The derivative of a function evaluated at a given point is the slope of the line tangent to the graph of the function at that point.

So, You've got a derivative, and you've got a point. Evaluate the derivative at that point and that will give the slope.

i.e. Plug in wherever you see an x, and whereever you see a y.

Note* .