Implicit differentiation with exponential function

[coolbeans33]

find dy/dx: e^xy+x²+y²= 5 at point (2,0)

I'm confused with finding the derivative with respect to x of e^xy.

this is what I did so far for just this part: e^xy*d(xy)/dx

e^xy*(y+x*dy/dx)

do I need to put the parentheses on here? I thought so because that is the part where I used the product rule. (but probably not, right?)

then for the entire function so far this is what I got:

e^xy*(y+x*dy/dx)+2x+2y*dy/dx=5

am I doing something wrong so far?

 

[MarkFL]

find dy/dx: e^xy+x²+y²= 5 at point (2,0)

I'm confused with finding the derivative with respect to x of e^xy.

this is what I did so far for just this part: e^xy*d(xy)/dx

e^xy*(y+x*dy/dx)

do I need to put the parentheses on here? I thought so because that is the part where I used the product rule. (but probably not, right?)

then for the entire function so far this is what I got:

e^xy*(y+x*dy/dx)+2x+2y*dy/dx=5

am I doing something wrong so far?

Your differentiation of the left side looks good (yes you do need the parentheses as given by the chain rule), but the right side is a constant, so after implicitly differentiating with respect to $x$, what should it become?

 

[coolbeans33]

Your differentiation of the left side looks good (yes you do need the parentheses as given by the chain rule), but the right side is a constant, so after implicitly differentiating with respect to $x$, what should it become?

srry I forgot to put the "equals zero" in.

So I solved for the derivative and I got:

ye^xy+2x/x*e^xy+2y

and the slope of the tangent line at pt (2,0) is 2.

 

[MarkFL]

We are given the implicit relation:

\(\displaystyle e^{xy}+x^2+y^2=5\)

Implicitly differentiating with respect to $x$, we find:

\(\displaystyle e^{xy}\left(x\frac{dy}{dx}+y \right)+2x+2y\frac{dy}{dx}=0\)

Next, we want to arrange this equation such that all terms having \(\displaystyle \frac{dy}{dx}\) as a factor are on one side, and the rest is on the other side:

\(\displaystyle xe^{xy}\frac{dy}{dx}+2y\frac{dy}{dx}=-\left(ye^{xy}+2x \right)\)

Factor the left side:

\(\displaystyle \frac{dy}{dx}\left(xe^{xy}+2y \right)=-\left(ye^{xy}+2x \right)\)

Divide through by \(\displaystyle xe^{xy}+2y\)

\(\displaystyle \frac{dy}{dx}=-\frac{ye^{xy}+2x}{xe^{xy}+2y}\)

Hence:

\(\displaystyle \left.\frac{dy}{dx} \right|_{(x,y)=(2,0)}=-\frac{0\cdot e^{2\cdot0}+2\cdot2}{2\cdot e^{2\cdot0}+2\cdot0}=-\frac{4}{2}=-2\)

Here is a plot of the equation and its tangent line:

View attachment 1603