#### Related to [[Derivative|derivatives]] --- ## Definition - An implicit function is a function that has **both x and y on the same side of the equation**. Therefore, implicit differentiation describes the methods that we use to **differentiate an implicit function**. --- ## Finding the derivative - Finding the derivative of an implicit function is *very similar* to finding the derivative of a regular function since all the **differentiation [[Derivative|rules]] are the same!**. - The only major difference is the appearance of x and y on the same side of the equation. As a result its important **to remember that the derivative of y is y' prime *not* 1 like the derivative of *x***. $\text{The derivative of y}=y'$ - After finding the derivative of the implicit function **use algebra to solve for y' prime** - the derivative of your implicit function. *This means that y' prime will be alone on one side of the equation.* >[!example] >$\text{\color{Bisque}\Large Example Problems}$ >$\newline$ >$\text{\color{Bisque}Problem 1:}$ >$(2x+3y^2)'\longrightarrow \left(2+6y\cdot \frac{dy}{dx}\right)\longrightarrow\color{Bisque}\left(\frac{dy}{dx}=\frac{-2}{6y}\right)$ ></br> >$\text{\color{Bisque}Problem 2:}$ >$\left( 2y\cdot x^2\right)'\longrightarrow\textit{\small Product rule:}\left(2\cdot\frac{dy}{dx}\cdot x^2+2y\cdot 2x\right)\longrightarrow\color{Bisque}\left(\frac{-2y\cdot 2x}{2x^2}\right)$ --- ## Constraints - A **constraint** is the term used to describe an equation **that solves for one of two unknown variables**. By plugging a constraint **back into its parent function** you can solve equations that contain **two unknown variables** *(like implicit derivatives!)*. - Constraints allow you to solve for **the x and y terms in an implicit derivative** only if you already know the result *(or slope)* of that derivative. ##### *Finding and using a constraint* - Once you have found the derivative of your implicit function you must also determine what you **want it to be equal to**. If you are, for example, trying to find the function's [[Derivative|critical points]] you want to find values of *x and y* when the derivative is equal to zero. >[!quote] $\text{\Large Steps for finding and using a constraint}$ </br> $\text{\color{LightCyan}Step 1 - Set }\space\color{LightGrey}\frac{dy}{dx}\space\text{ equal to the desired value:}$ $\left(\frac{dy}{dx}=\frac{2x-y}{y^2-3x}\right)\Rrightarrow\large\color{LightCyan}\left( 0=\frac{2x-y}{y^2-3x}\right)$ </br> $\text{\color{LightCyan}Step 2 - Solve }\text{\color{LightGrey} for one of your unknown variables:}$ $\left( 0=\frac{2x-y}{y^2-3x}\right) \rightarrow\Big(0=2x-y\Big)\Rrightarrow\large\color{LightCyan}\Big( y=2x\Big)$ </br> $\text{\color{LightCyan}Step 3 - Replace }\text{\color{LightGrey} x or y with the constraint and solve:}$ $\textit{\small\color{grey}Use the original function}\rightarrow\Big(5x+3y=1\Big)\rightarrow\Big(5x+3(2x)=1\Big)$ $\large\color{LightCyan}\therefore\left(x=\frac{1}{11}\right)$ $\newline$ >--- > $\textit{\small\color{grey}NOTE: The "original equation" shown in step 3 is technically incorrect since}$ $\textit{\small\color{grey}its derivative is different from the one above.}$ --- #mainpage