Second Derivatives of Implicit Functions (College Board AP® Calculus AB): Revision Note

Remember that:

• The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$

• The derivative of $\frac{dy}{dx}$ with respect to $x$ is $\frac{d^{2}y}{d{x}^2}$

Show that the second derivative of $\sin x+\cos y=0.8$ can be written as:

$\frac{d^{2}y}{d{x}^2}=-\frac{\sin x{\sin }^{2}y+{\cos }^{2}x\cos y}{{\sin }^{3}y}$

Answer:

Find the first derivative by differentiating both sides with respect to $x$

$\begin{array}{rcl}\frac{d}{dx}\left(\sin x+\cos y\right)& =& \frac{d}{dx}\left(0.8\right)\\ \frac{d}{dx}\sin x+\frac{d}{dx}\cos y& =& \frac{d}{dx}\left(0.8\right)\end{array}$

Remember to use the chain rule when differentiating $\cos y$ with respect to $x$

$\begin{array}{rcl}\cos x+-\sin y·\frac{dy}{dx}& =& 0\\ \frac{dy}{dx}& =& \frac{\cos x}{\sin y}\end{array}$

This could instead be simplified to $\cos x\csc y$, depending on whether you would prefer to apply the quotient rule to $\frac{\cos x}{\sin y}$ or the product rule to $\cos x\csc y$ for the next part

To find the second derivative, differentiate both sides with respect to $x$ again

$\begin{array}{rcl}\frac{d}{dx}\left(\frac{dy}{dx}\right)& =& \frac{d}{dx}\left(\frac{\cos x}{\sin y}\right)\\ \frac{d^{2}y}{d{x}^2}& =& \frac{d}{dx}\left(\frac{\cos x}{\sin y}\right)\\ & & \end{array}$

Apply the quotient rule to the right hand side, remember to apply the chain rule when differentiating $\sin y$ with respect to $x$

${\left(\frac{u}{v}\right)}^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

$u=\cos x$ $v=\sin y$

$u^{\text{'}}=-\sin x$ $v^{\text{'}}=\cos y·\frac{dy}{dx}$

$\frac{d^{2}y}{d{x}^2}=\frac{-\sin x·\sin y-\cos x·\cos y·{\displaystyle \frac{dy}{dx}}}{{\sin }^{2}y}$

Substitute in the expression for $\frac{dy}{dx}$ found previously, $\begin{array}{rcl}\frac{dy}{dx}& =& \frac{\cos x}{\sin y}\end{array}$

$\frac{d^{2}y}{d{x}^2}=\frac{-\sin x·\sin y-\cos x·\cos y·{\displaystyle \left(\frac{{\displaystyle \cos x}}{\sin y}\right)}}{{\sin }^{2}y}$

Now we need to rearrange into the form given in the question

Write as two fractions to see if they can be simplified

$\frac{d^{2}y}{d{x}^2}=\frac{-\sin x\sin y}{{\sin }^{2}y}-\frac{{\cos }^{2}x\cos y}{{\sin }^{3}y}$

They can be written with a common denominator

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{-\sin x{\sin }^{2}y}{{\sin }^{3}y}-\frac{{\cos }^{2}x\cos y}{{\sin }^{3}y} \\ \frac{d^{2}y}{d{x}^2}=\frac{-\sin x{\sin }^{2}y-{\cos }^{2}x\cos y}{{\sin }^{3}y} \end{gathered}$$

Factor -1 from the numerator to match the form given in the question

$\frac{d^{2}y}{d{x}^2}=-\frac{\sin x{\sin }^{2}y+{\cos }^{2}x\cos y}{{\sin }^{3}y}$