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implicit differentiation with exponential function

coolbeans33

New member
Sep 19, 2013
23
find dy/dx: exy+x2+y2= 5 at point (2,0)

I'm confused with finding the derivative with respect to x of exy.

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

exy*(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:

exy*(y+x*dy/dx)+2x+2y*dy/dx=5

am I doing something wrong so far?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
find dy/dx: exy+x2+y2= 5 at point (2,0)

I'm confused with finding the derivative with respect to x of exy.

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

exy*(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:

exy*(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

New member
Sep 19, 2013
23
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:

yexy+2x/x*exy+2y

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

coolbeans33.jpg