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.




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