What is the slope of the tangent line of # (x-y^2)/(xe^(x-y^2)) =C #, where C is an arbitrary constant, at #(1,1)#?

Question

Emily Ammerman · Accepted Answer

It is #1/2#

Given that #(1,1)# lies on the graph of # (x-y^2)/(xe^(x-y^2)) =C #,

we see that #C=0#

(Substitute #1# for both #x# and #y# to get #(1-(1)^2)/(1e^(1-(1^2))=0/1=0=C#.)

That means the graph is the graph of #x-y^2=0# or #y^2=x#.

Now we can differentiate implicitly, of further recognize that,

since #(1,1)# is on the graph, we are on the branch where #y=sqrtx#, so #y'=1/(2sqrtx)#

The slope at #x=1# is #1/2#.

Ezekiel Alexander · Answer

undefined To find the slope of the tangent line at the point (1,1) for the curve given by $ \frac{{x - y^2}}{{xe^{x - y^2}}} = C $, where C is an arbitrary constant, first, differentiate implicitly with respect to x to find $ \frac{{dy}}{{dx}} $, then evaluate it at the given point.

1. Differentiate $ \frac{{x - y^2}}{{xe^{x - y^2}}} = C $ implicitly with respect to x.
2. Solve for $ \frac{{dy}}{{dx}} $.
3. Substitute the values x = 1, y = 1 into $ \frac{{dy}}{{dx}} $ to find the slope of the tangent line at the point (1,1).