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Wize AP Calculus (BC) Textbook > Differentiation 2: Composite, Implicit, & Inverse Functions

Implicit Differentiation

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Derivatives of Inverse FunctionsNext Section
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Implicit Differentiation

We call a traditional function y=f(x)y=f(x) explicit since yy is in terms of xx. However, even if yy cannot be written explicitly in terms of xx, we can still compute its derivative by differentiating both sides of the equation.


Watch Out!
You must use implicit differentiation to take the derivative of equations that cannot be solved for yy explicitly (or would be very challenging to do so).

Procedure for Implicit Differentiation

  1. Differentiate on both sides of the equation by considering yy as a function of x.x.
  2. Use the chain rule for all yy dependent terms.
  3. Isolate dydx.\dfrac{dy}{dx}. (or y′y'depending on your preferred notation)

Wize Tip
Whenever you take the derivative of a yy , you must multiply the term by dydx\displaystyle \frac{dy}{dx} (or y′y'depending on your notation) due to the chain rule.
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Example: Implicit Differentiation

Find dydx\displaystyle \frac{dy}{dx} for the following equation:

ycos⁡x=x2+cos⁡yy\cos x=x^2+\cos y

Since we can't isolate for yy, we need to use implicit differentiation to find the derivative:
dydxcos⁡(x)+y(−sin⁡(x))=2x−sin⁡(y)dydx\begin{array}{rcl} \displaystyle \dfrac{dy}{dx}\cos(x)+y(-\sin(x)) &=& 2x-\sin(y)\dfrac{dy}{dx} \end{array}

Now solve for dydx\frac{dy}{dx}:
dydx(cos⁡(x)+sin⁡(y))=2x+ysin⁡(x)dydx=2x+ysin⁡(x)cos⁡(x)+sin⁡(y)\begin{array}{rcl} \displaystyle \dfrac{dy}{dx}(\cos(x)+\sin(y))& =& 2x+y\sin(x)\\ \displaystyle \dfrac{dy}{dx} &=& \dfrac{2x+y\sin(x)}{\cos(x)+\sin(y)} \end{array}
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Example: Implicit Differentiation (with 2nd Derivatives)

Find the second derivative, d2ydx2 \displaystyle \frac{d^2y}{dx^2} for the the implicit function:y2x=10\displaystyle \frac{y^2}{x}=10

First we will need to find the first derivative dydx\dfrac{dy}{dx}. (you may also use the y′y'notation)
y2x=10⇒x(2y×dydx)−y2(1)x2=0Note in the original expression, x≠0⇒2xy×dydx−y2=0⇒2xy×dydx=y2⇒dydx=y2x\begin{array}{l} \displaystyle \frac{y^2}{x}=10 \\ \\ \Rightarrow \displaystyle \frac{x(2y\times\dfrac{dy}{dx})-y^2(1)}{x^2}=0 \\ \\ \text{Note in the original expression, } x\ne0 \\ \\ \Rightarrow 2xy \times \dfrac{dy}{dx}-y^2=0 \\ \\ \Rightarrow 2xy \times \dfrac{dy}{dx}=y^2 \\ \\ \Rightarrow \dfrac{dy}{dx}=\dfrac{y}{2x} \end{array}

Now to find the second derivative, take the derivative of dydx\dfrac{dy}{dx}.

dydx=y2x⇒d2ydx2=2x(dydx)−y(2)4x2substitute: dydx=y2x⇒d2ydx2=2x(y2x)−2y4x2d2ydx2=y−2y4x2d2ydx2=−y4x2\begin{array}{l} \dfrac{dy}{dx}=\dfrac{y}{2x} \\ \\ \Rightarrow \displaystyle \frac{d^2y}{dx^2} = \dfrac{2x\left(\dfrac{dy}{dx}\right)-y(2)}{4x^2} \\ \\ \text{substitute: } \dfrac{dy}{dx}=\dfrac{y}{2x} \\ \\ \Rightarrow \displaystyle \frac{d^2y}{dx^2} = \dfrac{2x\left(\dfrac{y}{2x}\right)-2y}{4x^2} \\ \\ \displaystyle \frac{d^2y}{dx^2} = \dfrac{y-2y}{4x^2} \\ \\ \displaystyle \boxed{\frac{d^2y}{dx^2} = -\dfrac{y}{4x^2}} \end{array}

Find dydx\displaystyle\frac{dy}{dx} for the following equation:

tan⁡(yx)=x\displaystyle\tan\left(\frac{y}{x}\right)=x