Implicit differentiation is a technique that is used to determine the derivative of a function in the form y = f(x).

To learn how to use implicit differentiation, we can use the method on a simple example and then explore some more complex cases.

While it sounds more complicated, implicit differentiation uses all of the same mathematics and skills as basic differentiation. The important thing to note, however, is that our dependent variable now shows up in the function itself.

Take a simple equation such as xy = 1. There are two ways to find the derivative of y with respect to x, or dy/dx. First, we can simply solve for y in the equation and use the power rule for derivatives. Doing this would yield: y = 1/x. Applying the power rule would therefore reveal that dy/dx = -1/x².

We can also do this problem using implicit differentiation. Luckily, we already know the answer (it should be the same regardless of how we compute it), so we can check our work!

To begin, apply the derivative to both sides of the equation xy = 1. Then, d/dx(xy) = d/dx(1); clearly the right side is now equal to 0, but the left side requires the chain rule. This is because we are taking the derivative of our function, y, while it is being multiplied to another factor of x. To calculate this: d/dx(x)y + x(d/dx(y)) = y + xy'. We will use the prime notation to indicate a derivative with respect to x.

Rewriting our equation yields: y + xy' = 0. It's time to solve for y' in our equation! Clearly, y' = -y/x. But using the original information, we know that y= 1/x, so we can substitute this back in. Once we do that, we see that y' = -1/x², just like we found before.

To determine the derivative of y = sin(xy), we will use implicit differentiation by remembering that (d/dx)y = y'.

First, apply the derivative to both sides of the equation: d/dx(y) = d/dx (sin(xy)). The left side of the equation is clearly y', which is what we will need to solve for, but the right side will require some work; specifically, the chain rule and the product rule. First, the chain rule needs to be applied to sin(xy), and then the product rule for the argument xy. Luckily we calculated this product rule already.

Next, simplifying this gives: y' = cos(xy)(y + xy').

Clearly, this equation needs to be solved for y' in order to determine how y' is related to x and y.

Isolate all terms with y' on one side: y' - xy'cos(xy) = ycos(xy).

Then factor out the y' to get: y'(1 - xcos(xy)) = ycos(xy).

Now we see that y' = ycos(xy)/(1-xcos(xy)).

Further simplification my be necessary, but because our function is recursively defined, plugging in y = sin(xy) will likely not yield a satisfying solution. In this case, more information or a more sophisticated method for plotting these equations may be useful.

First, remember that implicit differentiation relies on one of the variables being a function of the other. Commonly, we see functions as y = f(x), but one could write a function x = f(y). Be careful when approaching these problems to determine which variable is dependent on the other.

Next, remember to carefully apply derivative rules. Implicit differentiation will require the chain rule very often, as well as the product rule and the quotient rule. Correctly applying these methods will be essential to determine the final answer.

Finally, solve for the desired derivative by isolating it, and simplifying the expressions as much as possible.