Implicit Partial Differentiation

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Implicit Partial Differentiation Implicit Partial Differentiation Consider the folium x3 + y3 – 9xy = 0 from Lesson 13.1. How would you find the slope of this curve at a given point? As before, the derivative will be used to find slope.

Differentiating Implicitly One method to find the slope is to take the derivative of both sides of the equation with respect to x. When taking the derivative of an expression that contains y, you must treat y as a function of x. This method is called implicit differentiation and it is illustrated below. Implicit differentiation of the folium x3 + y3 – 9xy = 0 yields . Notice that the term -9xy is considered a product of two functions and the Product Rule is used to find its derivative, . Solving for Know More About How To Simplify Trig Expressions

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Solving the expression for gives the derivative: Notice that the derivative contains both x and y. This is typical for implicit differentiation. Using the TI-89 to Differentiate Implicitly Another method used to find calculus:

is with your TI-89 and the following theorem from

If z is a differentiable function of x and y such that z(x, y) = 0 defines y implicitly as a differentiable function of x, then at any point where The symbol is called the partial derivative of z with respect to x. It means to take the derivative of z with respect to x while treating y as a constant. Similarly, is the partial derivative of z with respect to y. To find this partial derivative, take the derivative of z with respect to y while treating x as a constant. Finding Partial Derivatives With z1= x3 + y3 – 9xy,

can be found on the TI-89 with the derivative command.

Notice that parentheses are used to indicate the variables x and y of the function z1. Enter -d(z1(x,y),x)/d(z1(x,y),y) Read More About Rounding Fractions To The Nearest Whole Number

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Solving Nonlinear Systems of Equations You can find the coordinates of the points on the folium in this case by finding the solutions to a nonlinear system of equations where one equation defines the set of points on the graph and a second equation defines the set of points where the derivative is zero. Solutions to the system below are the points on x3 + y3 – 9xy = 0 where the tangent may be horizontal. x3 + y3 – 9xy = 0 and 3y – x2 = 0 The solutions to the system can be found by using the Solve command to find values for the two variables x and y that make both equations true. Notice that parentheses are used to indicate that z1(x, y) is a function of x and y and that the variables to be found, x and y, are enclosed within braces. solve(z1(x,y)=0 and 3y-x^2=0,{x,y})

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Thank You

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