Question

Could really use a lot of help here

what is the slope of the tangent line to
cos(x-y)=xe^x at point (0,3pi/2)

Answer 1

how so? It's just a straightforward implicit differentiation problem. (with a little chain rule and product rule thrown in)

-sin(x-y)(1 - y') = e^x + xe^x
1-y' = e^x(x+1)/-sin(x-y)
y' = 1 + e^x(x+1)/sin(x-y)

So, at (0,3pi/2),
y' = 1 + 1(0+1)/sin(0-3pi/2)
= 1 + 1/1
= 2