# Differentials

Standard

In almost all of the examples and applications so far, all functions have been defined in the form y=f (x) with y given explicitly in terms of x, or as an explicit function of x, for example y = 3x – 2 is an all explicit functions of x. The equation 4xy – 3x = 6 can be expressed as an explicit function of x by solving for y. This is:

4xy – 3x= 6

4xy = 3x + 6

y = $\frac {3x+6}{4x}$

On the other hand, some equations in x and y cannot be readily solved for y, and some equations cannot be solved for y at all. For example, while it would be possible (but tedious) to use the formula to solve for y in the equation $y^2+2yx+4x^2=0$, it is not possible to solve for y in the equation $y^5+by^3+6y^2x^2+2yx^3+6=0.$ In equations such as these last two, y is said to be given implicitly in terms of x.

In such cases, it may still be possible to find the derivative $\frac {dy}{dx}$  by a process called implicit differentiation. In doing so, we assume that there exists some function or functions f, which we may not be able to find, such that y= f(x) and $\frac {dy}{dx}$ exsist. It is useful to use $\frac {dy}{dx}$  here rather than f'(x) to make it clear which variable is independent and which is dependent.

One notation for the derivative of y is with respect to x is $\frac {dy}{dx}$ Notice that this similar to one way of starting slope which is $\frac {\Delta y}{\Delta x}$. $\Delta$ is used to represent change. When considering $\frac {dy}{dx}$ it is often useful to tink of the x as an infinitesimally small change. Then dy is the corresponding change in y.

REVIEW: Earlier in the year we pointed out that when y is given as a function of x, x is the independent variable. We later defined the derivative $\frac {dy}{dx}$ when y is a function of x. in an equation such as  3xy + $4y^2$ = 10 , either variable can be considered the independent variable; if it asks for $\frac {dy}{dx}$, consider x the independent variable ; if it ask for $\frac {dx}{dy}$, consider y the independent variable. A similar rule holds when other variables are used.

If y= f(x) then $\frac {dy}{dx}$ = f”(x); dy= f'(x)dx

Steps to using Implicit Differentiation:

To find $\frac {dy}{dx}$ for an equation containing x and y:

1. Differentiate on both sides of the equation with respect to x, keeping in mind that y is assumed to be a function of x.

2. Place all terms with $\frac {dy}{dx}$ on one side of the equal sign, and al terms without $\frac {dy}{dx}$ on the other side.

3. Factor out $\frac {dy}{dx}$, and then solve for $\frac {dy}{dx}$ .

HINT: When an applied problem involves an equation that is not given in explicit form, implicit differentiation can be used to locate maxima and minima or to find rates of change.

GOOD LUCK ON THE TEST MONDAY! 😀