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

Question

Emilia Aguilar · Accepted Answer

#m = 1/10#

Given that the curve contains the point #(1,1)#, we find #C = 9#

The equation is equivalent to

#x^2+4xy+4y^2 = 9e^(x-y^2)#.

Differentiate implicitly to get

#2x+4y+4xdy/dx+8ydy/dx=9e^(x-y^2)(1-2ydy/dx)#

At #(1,1)# we have

#2+4+4dy/dx+8dy/dx=9-18dy/dx#

so, #dy/dx = 1/10#

Scarlett Allison · Answer

undefined To find the slope of the tangent line at the point (1, 1) for the curve given by $\frac{(x+2y)^2}{e^{(x-y^2)}} = C$, we need to find the derivative of $y$ with respect to $x$ implicitly using the implicit differentiation method.

Implicitly differentiating the given equation with respect to $x$, we get:

$$
\frac{d}{dx}\left(\frac{(x+2y)^2}{e^{(x-y^2)}}\right) = 0
$$

Now, using the quotient rule and chain rule for differentiation, we find:

$$
\frac{d}{dx}\left(\frac{(x+2y)^2}{e^{(x-y^2)}}\right) = \frac{2(x+2y)(1+2y')e^{(x-y^2)} - (x+2y)^2e^{(x-y^2)}(1-2y^2)}{e^{2(x-y^2)}}
$$

Given that the point is (1, 1), substitute $x = 1$ and $y = 1$ into the derivative expression.

After simplification, evaluate the derivative expression at (1, 1) to find the slope of the tangent line at that point.