Implicit Differentiation (College Board AP® Calculus AB): Revision Note

Derivatives of implicit functions

What is an implicit function?

• An equation in the form $y=f\left(x\right)$ or $x=f\left(y\right)$is said to be written explicitly

  • E.g. $y=3x^2+2x-3$

• Equations in terms of both $x$ and $y$ which cannot be written as $y$ in terms of $x$ are referred to as implicit functions

  • E.g. $3x^2-7x{y}^2=3$ or $x^2+y^2=25$

  • For such an equation

    • We cannot write the relationship between $x$ and $y$ explicitly by expressing $y$ as a function of $x$

    • But values of $y$ satisfying the equation still depend implicitly on the values of $x$

What is implicit differentiation?

• The method by which implicit functions are differentiated is known as implicit differentiation

• To differentiate an implicit function with respect to $x$, each term is differentiated with respect to $x$

• For terms only involving $x$, this is straightforward

• However, if an expression is in terms of $y$, then to differentiate it with respect to $x$, we apply the chain rule

  • $\frac{d}{dx}f\left(y\right)=f^{\text{'}}\left(y\right)·y^{\text{'}}=f^{\text{'}}\left(y\right)·\frac{dy}{dx}$

  • In short, this means:

    • Differentiate the function that is in terms of $y$, with respect to $\mathit{y}$,

    • and then multiply it by the term $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$

• This could also be written as

  • $\frac{du}{dx}=\frac{du}{dy}·\frac{dy}{dx}$ where $u$ is a function in terms of $y$

• Once each term has been differentiated with respect to $x$,

  • rearrange to make $\frac{dy}{dx}$ the subject

  • you may have to factor out $\frac{dy}{dx}$ to do this

• The final expression for $\frac{dy}{dx}$ will often be in terms of both $x$ and $y$

  • To then find the value of the derivative at a point, you would need the coordinates of a point $\left(x,y\right)$ to substitute in

    • And substitute in both the $x$ and $y$ values

  • This is only valid if the point $\left(x,y\right)$ lies on the curve in question

    • To check this, substitute the $x$ and $y$ values into the original equation of the curve, and check that the equation is satisfied

      • If it is then the point lies on the curve

      • If it is not, then the point does not lie on the curve

How do I use implicit differentiation?

• For example, to differentiate $x^2+y^2=4x$

• The derivative would be $\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(4x\right)$

• The terms that are only in terms of $x$ are straight forward

  • $2x+\frac{d}{dx}\left(y^2\right)=4$

• Then apply the result $\frac{d}{dx}f\left(y\right)=f^{\text{'}}\left(y\right)·\frac{dy}{dx}$ to the remaining term, which is in terms of $y$

  • $2x+\left(2y·\frac{dy}{dx}\right)=4$

• Rearrange to find an expression for $\frac{dy}{dx}$

  • $\frac{dy}{dx}=\frac{4-2x}{2y}=\frac{2-x}{y}$

How can I use the chain rule and product rule with implicit differentiation?

• Implicit differentiation may be combined with other skills including

  • Chain rule

  • Product rule

  • Quotient rule

• If a term involves $y$ then remember to multiply its derivative by $\frac{dy}{dx}$

  • For example, consider $\frac{d}{dx}\left(xy\right)$

    • The product rule gives $\frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(x\right)·y+x\frac{d}{dx}\left(y\right)$

    • Which simplifies to $\frac{d}{dx}\left(xy\right)=y+x\frac{dy}{dx}$

  • For example, consider $\frac{d}{dx}\left(\sin \left(x+y\right)\right)$

    • The chain rule gives $\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)·\frac{d}{dx}\left(x+y\right)$

    • Which simplifies to $\frac{d}{dx}\left(\sin \left(x+y\right)\right)=\cos \left(x+y\right)·\left(1+\frac{dy}{dx}\right)$

Derivatives of inverse functions using implicit differentiation

How can I differentiate inverse functions using implicit differentiation?

• Implicit differentiation provides an alternative method for finding the derivative of inverse functions

• Consider differentiating $y={\sin }^{-1}x$

  • This can be rewritten as $x=\sin y$

• Differentiate using implicit differentiation

  • $\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(\sin y\right)$

  • $1=\cos y·\frac{dy}{dx}$

• Rearrange to make $\frac{dy}{dx}$ the subject

  • $\frac{dy}{dx}=\frac{1}{\cos y}$

• Use the identity ${\sin }^{2}y+{\cos }^{2}y=1$, rearranged to make $\cos y$ the subject

  • ${\cos }^{2}y=1-{\sin }^{2}y$

  • $\cos y=\pm \sqrt{1-{\sin }^{2}y}$

• Therefore

  • $\frac{dy}{dx}=\pm \frac{1}{\sqrt{1-{\sin }^{2}y}}$

• But $x=\sin y$, so

  • $\frac{dy}{dx}=\pm \frac{1}{\sqrt{1-x^2}}$

• A similar method can be used to show why the derivative of $a^x$ is $a^x\ln a$

  • If $y=a^x$

    • then $\ln y=x\ln a$

  • Differentiate using implicit differentiation

    • Remember that $\frac{d}{dx}\left(\ln x\right)=\frac{1}{x}$ and that $\ln a$ is a constant

    • $\frac{1}{y}·\frac{dy}{dx}=\ln a$

  • Rearrange

    • $\frac{dy}{dx}=y\ln a$

  • Recall that $y=a^x$

    • $\frac{dy}{dx}=a^x\ln a$