This Article is about finding the derivative of y in respect to x, when y cannot be written explicitly in terms of x alone. So to find the derivative of y in respect to x we need to do so by Implicit differentiation. This Article will show how this is done.

Given the Equation y = sin(xy), we will show how to do the Implicit differentiation of this equation by two different methods. The first method is differentiating by finding the derivative of the x-terms as we usually do and using the Chain Rule when differentiating the y-terms. Please click on the Image for a better understanding.

We will now take this differential equation, dy/dx = [x(dy/dx) + y(1)]cos(xy), and solve for dy/dx. that is, dy/dx = x(dy/dx)cos(xy) + ycos(xy), we distributed the cos(xy) term. We will now collect all dy/dx terms on the left side of the equal sign. (dy/dx)- xcos(xy)(dy/dx) = ycos(xy). By factoring out the (dy/dx) term, 1 - xcos(xy) = ycos(xy), and solving for dy/dx, we get.... dy/dx = [ycos(xy)]/[1 - xcos(xy)]. Please click on the Image for a better understanding.

## Sciencing Video Vault

The second method of differentiating the Equation y = sin(xy), is differentiating the y-terms in respect to y and the x-terms in respect to x, then dividing each term of the equivalent equation by dx. Please click on the Image for a better understanding.

We will now take this differential equation, dy = [xdy + ydx]cos(xy) and distribute the cos(xy) term. That is, dy = xcos(xy)dy + ycos(xy)dx, we now divide each term of the equation by dx. We now have, (dy/dx) = [xcos(xy)dy]/dx + [ycos(xy)dx]/dx, which is equal to...dy/dx = xcos(xy) + ycos(xy). Which is equivalent to, dy/dx =xcos(xy) + ycos(xy). To solve for dy/dx, we go to step #2. That is We will now collect all dy/dx terms on the left side of the equal sign. (dy/dx)- xcos(xy)(dy/dx) = ycos(xy). By factoring out the (dy/dx) term, 1 - xcos(xy) = ycos(xy), and solving for dy/dx, we get.... dy/dx = [ycos(xy)]/[1 - xcos(xy)]. Please click on the Image for a better understanding.