y is an explicity function of x if it is written y= f(x) for some expression f-. That is, we have y by itself on one side of the equation, an expression that involves only x on the other side. In particular, that means that if we are given a value of f, we only have to do the operations indicated in the formula- we are given explicit instructions on how to find the values. Examples would be \(\displaystyle y= x^2\), \(\displaystyle y= sin(x)\), and \(\displaystyle y= 3^x\).
y is an implicit function of x if we are given an equation in which both x and y occur. If we are given a value of x, we could put it into the equation and then have to solve the equation for y- we are implicitely told how to find the values. Examples would be \(\displaystyle y^3- x= 0\), \(\displaystyle sin(xy)= 1/2\), and \(\displaystyle x^2- 2xy+ y^2= 4\).
As for finding the derivative, the basic idea is really the same- with the obvious changes. To differentiate \(\displaystyle y= x^2\) we take the derivative of both sides with respect to x- of course, on the left we just write the derivative "symbol": \(\displaystyle y'= 2x\). If we have an "implicit" function \(\displaystyle x^2- 2xy+ y^2= 4\), differentiating both sides, with respect to x, again writing "y'" for the derivative of y.- \(\displaystyle (x^2)'- (2xy)'+ (y^2)'= (4)'\), \(\displaystyle (2x)- (2y+ 2xy')+ 2yy'= 0\), using the product rule for \(\displaystyle (2xy)'\) and the chain rule for \(\displaystyle (y^2)'\), and then solve for y'. We can write that as \(\displaystyle 2(x- y)+ 2(y+1)y'= 0\) and then \(\displaystyle (y+1)y'= y- x\) so that \(\displaystyle y'= \frac{y- x}{y+ 1}\).