# 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

${\displaystyle x^{5}+y^{5}-5xy+1=0}$

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

Since

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

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

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

In doing this we get

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

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

Thus giving us

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

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 ${\displaystyle \operatorname {d} y \over \operatorname {d} x}$ over to the other side. Giving us

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

Now we simplify and get

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

Other problems to work on

Ex. 2

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

${\displaystyle xy^{2}+x^{2}y=1}$

Ex. 3

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

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