How do you find the derivative of #7=xy-e^(xy)#?

Question

Charles Anctil · Accepted Answer

#dy/dx = -y/x#

By implicit differentiation:

#d/dx(7) = d/dx(xy - e^(xy))#

#d/dx(7) = d/dx(xy) + d/dx(e^(xy))#

By using the identity #d/dx(e^(b(x))) = (b(x))' xx e^(b(x))#

#0 = y + x(dy/dx) + (y + x(dy/dx))e^(xy)#

#0 = y + x(dy/dx) + ye^(xy) + xe^(xy)(dy/dx)#

#-y - ye^(xy) = dy/dx(x + xe^(xy))#

#(-y - ye^(xy))/(x + xe^(xy)) = dy/dx#

#-(y + ye^(xy))/(x + xe^(xy)) = dy/dx#

#dy/dx = -(y(1 + e^(xy)))/(x(1 + e^(xy)))#

#dy/dx = -y/x#

Hopefully this helps!

Emma Aldridge · Answer

undefined To find the derivative of $7 = xy - e^{xy}$, you would use implicit differentiation. Taking the derivative of each term with respect to $x$, you get:

$$\frac{d}{dx}(7) = \frac{d}{dx}(xy) - \frac{d}{dx}(e^{xy})$$

Solve each derivative separately:

$$\frac{d}{dx}(7) = 0$$

$$\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$$

$$\frac{d}{dx}(e^{xy}) = e^{xy}(y + x\frac{dy}{dx})$$

Substitute these derivatives back into the original equation and solve for $\frac{dy}{dx}$:

$$0 = y + x\frac{dy}{dx} - e^{xy}(y + x\frac{dy}{dx})$$

$$0 = y + x\frac{dy}{dx} - e^{xy}y - x e^{xy}\frac{dy}{dx}$$

$$0 = y(1 - e^{xy}) + x(\frac{dy}{dx} - e^{xy}\frac{dy}{dx})$$

$$0 = y(1 - e^{xy}) + x(1 - e^{xy})\frac{dy}{dx}$$

$$\frac{dy}{dx} = \frac{y(1 - e^{xy})}{x(e^{xy} - 1)}$$