High School Calculus/Implicit Differentiation

Implicit Differentiation
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 $$x+y+(x-y)^2+(2x-3y)^3=0$$