High School Calculus/Implicit Differentiation

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Implicit Differentiation[edit]

When a functional relation between x and y cannot be readily solved for y, the preceding rules may be applied directly to the implicit function.

The derivative will usually contain both x and y. Thus the derivative of an algebraic function, defined by setting the polynomial of x and y to zero.

Ex. 1

Given the function y of x

x^5+y^5-5xy+1=0

Find {\operatorname{d}y\over\operatorname{d}x}

Since

{\operatorname{d}\over\operatorname{d}x}(x^5+y^5-5xy+1)=0

=5x^4+5y^4{\operatorname{d}y\over\operatorname{d}x}-5y-5x{\operatorname{d}y\over\operatorname{d}x}=0

In solving for {\operatorname{d}y\over\operatorname{d}x} we must first factor the differentiation problem

In doing this we get

{\operatorname{d}y\over\operatorname{d}x}(5y^4-5x)+(5x^4-5y)=0

From here we subtract the {\operatorname{d}y\over\operatorname{d}x} to one side

Thus giving us

5x^4-5y=-{\operatorname{d}y\over\operatorname{d}x}(-5x+5y^4)

Here I am going to skip a step in solving this implicit differentiation problem. I am going to skip the step where I divide the -1 over to the other side.

From here we divide the polynomial from the \operatorname{d}y\over\operatorname{d}x over to the other side. Giving us

\left (\frac{-5x^4+5y}{-5x+5y^4}\right)={\operatorname{d}y\over\operatorname{d}x}

Now we simplify and get

{\operatorname{d}y\over\operatorname{d}x}=\left (\frac{x^4-y}{x-y^4}\right)

Other problems to work on

Ex. 2

Find {\operatorname{d}y\over\operatorname{d}x} given the function

xy^2+x^2y=1

Ex. 3

Find {\operatorname{d}y\over\operatorname{d}x} given the function

x+y+(x-y)^2+(2x-3y)^3=0