# 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

$xy^{2}+x^{2}y=1$ Ex. 3

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

$x+y+(x-y)^{2}+(2x-3y)^{3}=0$ 