Implicit Differentiation

Implicit Differentiation We'll look at a second way to compute derivatives, called implicit differentiation. We'veseen the derivatives of some well-known simple functions; however, sometimes functionsare defined in an complicated way and not necessarily written out as y equals a certainformula in terms of x. Example. Consider the equation y = x2 + 1. The derivative of this equation with respect to x is not a simple function of x. Or consider the equation y = x(1 + 2x), where we cannotsimply write out the derivative of y with respect to x as a function of x. In these situations,how do we find the derivative of y with respect to x? The way we will differentiate these equations is by using the implicit derivative. It is veryeasy to apply, as we will just try to differentiate both sides with respect to x and then solvefor dy/dx algebraically. Let us illustrate the method by means of an example. In this equation, Y is defined as afunction of x. We differentiate everything that appears in this equation with respect to x, sowe differentiate the left-hand side and obtain this term first, then differentiate this term,then differentiate this term.

On the right-hand side, differentiate zero. Of course, you get zero. Now differentiate thisterm here and you get 2y dy/dx. Differentiate this term and use the chain rule to get minusone over y - x2. Then differentiate y - x. So, differentiate y - x is dy/dx - dx/dx. To do this, use the product rule. You get xdy/dx + ydx/dx = y. Now you try to find out from this equation dy/dx. You regard dy/dx as an unknown quantity and then collect the terms involving dy/dx. Youthen move all other terms not involving dy/dx to the right side, leaving dy/dx alone on theleft side of the equation. This gives you the formula for dy/dx. The equation x equals sin y defines the function y in terms of x. In this case, the functionthat y in terms of x is y equals sin minus one x. We will find dy/dx by implicit differentiation. To find the derivative of sine, we start with the formula for differentiation and differentiateboth sides. The left-hand side is dx/dx, which equals one; the right-hand side getsdifferentiated and gives us cosine and then dy/dx.

Suppose we want to find dy/dx. We know that dy/dx is equal to one over cosine y. We canfind out what cosine y is by recalling a well-known formula, cosine y squared equals oneminus sin y squared. So, putting this value here, we get this expression for dy/dx. Now remember that y is equalto sin minus one x, so this formula also says differentiation of sin minus one x is equal tothis which is the same formula as we have seen earlier. We now look at the equation x2 + y2 = 9. We see that it defines a circle in the x-y plane,but it actually defines two functions. The circle in the x-y plane given by the equation x square plus y square equals nine iscentered at the origin and has a radius of three. It defines two functions: y one, square rootof nine minus x square; and x between minus three and three. The graph of this function is the blue one above the x-axis. It also defines another functionthat I denote by y cubed, which is minus square root of nine minus x square for the samerange of x. If the graph of this function is y two, the lower half of the circle represents a certainequation. However, this equation actually defines two functions: first, let's try to find thederivative for y one. To find the derivative of y two, we can use the formula dy = y dx. The result is dy = 2x - 1.To find the derivative of x one, we can use another formula, dx = x dy. Here, the result is dx= 2y - 1. The difference between these two formulas is a negative sign. We can differentiate this equation implicitly to find the derivative. Differentiating both sideswith respect to x, we get the following. On the left-hand side, we have: dx square, dx plus dy squared dx equals to d nine dxwhich is just zero. The derivative of this left-hand side equals two x plus two y, dy/dx. We can solve dy/dx as an unknown from this equation, and obtain dy/dx = -x/y. Now whythis is just one single formula while in the previous method we have two formulas?

They are the same because y can be either y one or y two. We get this formula here. If yequals to y two, then we get this formula here.