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

Revision Note

Author

Jamie Wood

Expertise

Maths

Second derivatives of implicit functions

How do I find the second derivative of an implicit function?

First make sure you are comfortable with the content covered in the 'Implicit Differentiation' study guide
- This involves being able to find the derivative of a function defined implicitly, using the chain rule
We are able to find the second derivative, $\frac{d^{2} y}{d x^{2}}$ , by differentiating the expression for the first derivative, $\frac{d y}{d x}$
Consider finding the second derivative of $x^{2} + y^{2} = 4 x$
Find the first derivative, this is shown in the Implicit Differentiation study guide to be
- $\frac{d y}{d x} = \frac{2 - x}{y}$
To find the second derivative, differentiate both sides with respect to $x$
- $\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{2 - x}{y})$
The derivative of $\frac{d y}{d x}$ with respect to $x$ is $\frac{d^{2} y}{d x^{2}}$
- $\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{2 - x}{y})$
The right hand side can then be differentiated with respect to $x$ , in this case using the quotient rule, ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$
- $u = 2 - x$ and $v = y$
Differentiate $u$ and $v$ with respect to $x$ , remember to apply the chain rule when differentiating $y$
- $u^{'} = - 1$ and $v^{'} = \frac{d y}{d x}$
Applying the quotient rule to the right hand side
- $\frac{d^{2} y}{d x^{2}} = \frac{- 1 \cdot y - (2 - x) \cdot \frac{d y}{d x}}{y^{2}}$
If your answer references the first derivative, substitute it in
- We know from before that $\frac{d y}{d x} = \frac{2 - x}{y}$
- $\frac{d^{2} y}{d x^{2}} = \frac{- 1 \cdot y - (2 - x) \cdot (\frac{2 - x}{y})}{y^{2}}$
This is the correct second derivative, but it can be simplified
- $\frac{d^{2} y}{d x^{2}} = \frac{- y - \frac{{(2 - x)}^{2}}{y}}{y^{2}}$
Writing with a common denominator can help simplify
- $\frac{d^{2} y}{d x^{2}} = \frac{- y}{y^{2}} - \frac{{(2 - x)}^{2}}{y^{3}} = \frac{- y^{2}}{y^{3}} - \frac{{(2 - x)}^{2}}{y^{3}}$
$\frac{d^{2} y}{d x^{2}} = - \frac{y^{2} + {(2 - x)}^{2}}{y^{3}}$
This is the final answer, but expressions like this can be written in several different ways,
- E.g. by factoring out negative signs in a different place

Exam Tip

Remember that:

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

Worked Example

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 $x$

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

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

$\begin{array}{rcl} \cos x + - \sin y \cdot \frac{d y}{d x} & = & 0 \\ \frac{d y}{d x} & = & \frac{\cos x}{\sin y} \end{array}$

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

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

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

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

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

$u^{'} = - \sin x$ $v^{'} = \cos y \cdot \frac{d y}{d x}$

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

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

$\frac{d^{2} y}{d x^{2}} = \frac{- \sin x \cdot \sin y - \cos x \cdot \cos y \cdot (\frac{\cos x}{\sin y})}{\sin^{2} y}$

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

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

Put the negative at the front of the entire fraction, to match the form given in the question

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

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