3.3 Higher order derivatives

Achievable AP Calculus AB

3. Advanced differentiation

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Higher order derivatives

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What you’ll learn

How to find the 2nd derivative with implicit differentiation

In the previous section, we found the first derivative of a curve by implicitly differentiating. The first derivative tells you the slope of the tangent line - how $y$ changes as $x$ changes - and whether the curve is increasing or decreasing as you move from left to right.

Differentiating the first derivative gives the second derivative, denoted $y^{''}$ or $\frac{d ^{2} y}{d x ^{2}}$ . The second derivative describes how the slope itself is changing, which helps reveal the shape of a curve.

For implicit equations, finding the 2nd derivative often requires implicitly differentiating twice.

Using implicit differentiation, the first derivative of a circle

$x^{2} + y^{2} = 25$

was found to be

$\frac{d y}{d x} = - \frac{x}{y}$

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

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (- \frac{x}{y})$

The left side becomes $\frac{d ^{2} y}{d x ^{2}}$ , the second derivative.

On the right side, use the quotient rule. Remember to use the chain rule when differentiating a term involving $y$ with respect to $x$ . For readability, the 1st derivative $\frac{d y}{d x}$ will be written as $y^{'}$ .

$\frac{d ^{2} y}{d x ^{2}} = - (\frac{y ( 1 ) - x ( y ^{'} )}{y ^{2}}) = - \frac{y - x y ^{'}}{y ^{2}}$

Since $y^{'} = - \frac{x}{y}$ , substitute this in:

$\frac{d ^{2} y}{d x ^{2}} = - \frac{y - x ( - \frac{x}{y} )}{y ^{2}} = - \frac{y + \frac{x ^{2}}{y}}{y ^{2}}$

Simplifying,

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

Because the original equation tells us $x^{2} + y^{2} = 25$ , we can also write:

$\frac{d ^{2} y}{d x ^{2}} = - \frac{25}{y ^{3}}$

Example

Evaluate $\frac{d ^{2} y}{d x ^{2}}$ at the point $(1, - 2)$ for

$2 x^{4} - x y = y^{2}$

Solution

Step 1: Find $\frac{d y}{d x}$

(spoiler)

Differentiating implicitly,

$8 x^{3} - (x \cdot \frac{d y}{d x} + y) = 2 y \cdot \frac{d y}{d x}$

Solve for $\frac{d y}{d x}$ by combining like terms:

$8 x^{3} - x \cdot \frac{d y}{d x} - y - x \cdot \frac{d y}{d x} - 2 y \cdot \frac{d y}{d x} - \frac{d y}{d x} (x + 2 y) \frac{d y}{d x} (x + 2 y) \frac{d y}{d x} = 2 y \cdot \frac{d y}{d x} = y - 8 x^{3} = y - 8 x^{3} = 8 x^{3} - y = \frac{8 x ^{3} - y}{x + 2 y}$

Step 2: Find $\frac{d ^{2} y}{d x ^{2}}$

(spoiler)

To find the 2nd derivative, implicitly differentiate again while applying the quotient rule. For readability, $\frac{d y}{d x}$ is written as $y^{'}$ .

$\frac{d ^{2} y}{d x ^{2}} = \frac{( x + 2 y ) ( 24 x ^{2} - y ^{'} ) - ( 8 x ^{3} - y ) ( 1 + 2 y ^{'} )}{( x + 2 y ) ^{2}}$

Step 3: Evaluate at the point

(spoiler)

To evaluate at $(1, - 2)$ , we also need the value of $y^{'}$ at the point.

$y^{'} = \frac{8 x ^{3} - y}{x + 2 y} = \frac{8 ( 1 ) ^{3} - ( - 2 )}{1 + 2 ( - 2 )} = \frac{8 + 2}{1 - 4} = - \frac{10}{3}$

Then the value of $\frac{d ^{2} y}{d x ^{2}}$ at $(1, - 2)$ equals

$\frac{d ^{2} y}{d x ^{2}} = \frac{( x + 2 y ) ( 24 x ^{2} - y ^{'} ) - ( 8 x ^{3} - y ) ( 1 + 2 y ^{'} )}{( x + 2 y ) ^{2}} = \frac{( 1 - 4 ) ( 24 + \frac{10}{3} ) - ( 8 + 2 ) ( 1 + 2 ( \frac{10}{3} ))}{( 1 - 4 ) ^{2}} = \frac{( - 3 ) ( \frac{82}{3} ) - ( 10 ) ( \frac{23}{3} )}{9} = - \frac{476}{27}$

Challenge problem

Consider the curve defined by the equation:

$a x^{2} + b x y + y^{2} = 12$

where $a$ and $b$ are constants.

a) If the tangent line to the curve at the point $(2, 4)$ has a slope of $1,$ find the values of $a$ and $b$ .

b) Evaluate $\frac{d ^{2} y}{d x ^{2}}$ at $(2, 4)$ .

Solutions

a) Find $a$ and $b$ .

(spoiler)

Differentiating implicitly (noting that $a$ and $b$ are constants):

$a (2 x) + b (x \cdot \frac{d y}{d x} + y) + 2 y \cdot \frac{d y}{d x} = 0$

Solving for $\frac{d y}{d x}$ ,

$2 a x + b x \cdot \frac{d y}{d x} + b y + 2 y \cdot \frac{d y}{d x} \frac{d y}{d x} (b x + 2 y) \frac{d y}{d x} = 0 = - 2 a x - b y = \frac{- 2 a x - b y}{b x + 2 y}$

If the tangent line at $(2, 4)$ has a slope of $1$ , then $\frac{d y}{d x} = 1$ when $x = 2$ and $y = 4$ . Substitute these values:

$11 2 b + 8 4 a + 6 b = \frac{- 2 a ( 2 ) - b ( 4 )}{b ( 2 ) + 2 ( 4 )} = \frac{- 4 a - 4 b}{2 b + 8} = - 4 a - 4 b = - 8$

This is one equation relating $a$ and $b$ . To get a second equation, use the equation of the original curve at $(2, 4)$ :

$a x^{2} + b x y + y^{2} a (2)^{2} + b (2) (4) + (4)^{2} 4 a + 8 b = 12 = 12 = - 4$

Solve the system of equations

$4 a + 6 b 4 a + 8 b = - 8 = - 4$

to obtain

$a = - 5 b = 2$

b) Evaluate $\frac{d ^{2} y}{d x ^{2}}$ at $(2, 4)$ .

(spoiler)

Since $a = - 5$ and $b = 2$ ,

$\frac{d y}{d x} = \frac{- 2 a x - b y}{b x + 2 y} = \frac{10 x - 2 y}{2 x - 10 y} = \frac{5 x - y}{x - 5 y}$

Then to find the 2nd derivative, implicitly differentiate:

$\frac{d ^{2} y}{d x ^{2}} = \frac{( x - 5 y ) ( 5 - y ^{'} ) - ( 5 x - y ) ( 1 - 5 y ^{'} )}{( x - 5 y ) ^{2}}$

At $(2, - 4)$ , the 1st derivative is

$y^{'} = \frac{5 ( 2 ) - ( - 4 )}{2 - 5 ( - 4 )} = \frac{10 + 4}{2 + 20} = \frac{14}{22} = \frac{7}{11}$

Then substituting into the 2nd derivative,

$\frac{d ^{2} y}{d x ^{2}} = \frac{( 2 - 5 ( - 4 )) ( 5 - \frac{7}{11} ) - ( 5 ( 2 ) - ( - 4 )) ( 1 - 5 ( \frac{7}{11} ))}{( 2 - 5 ( - 4 ) ) ^{2}} = \frac{( 22 ) ( \frac{48}{11} ) - ( 14 ) ( - \frac{24}{11} )}{2 2 ^{2}} = \frac{348}{1331}$

Second derivative via implicit differentiation

Second derivative = derivative of the first derivative; notation: $\frac{d ^{2} y}{d x ^{2}}$
Found by implicitly differentiating twice with respect to $x$
After second differentiation, substitute the expression for $y^{'}$ to simplify

Key technique: substituting $y^{'}$

After applying quotient/chain rule a second time, replace $y^{'}$ with the first-derivative expression
Substitute original equation constraints (e.g., $x^{2} + y^{2} = 25$ ) to further simplify result
Example result for circle: $\frac{d ^{2} y}{d x ^{2}} = - \frac{x ^{2} + y ^{2}}{y ^{3}} = - \frac{25}{y ^{3}}$

Evaluating at a point

Calculate $y^{'}$ at the given point first, then substitute into the $\frac{d ^{2} y}{d x ^{2}}$ expression
Both $x$ , $y$ , and $y^{'}$ values must be substituted simultaneously
Use original curve equation to find unknown constants before differentiating, if needed

Higher order derivatives

What you’ll learn

How to find the 2nd derivative with implicit differentiation

For implicit equations, finding the 2nd derivative often requires implicitly differentiating twice.

Using implicit differentiation, the first derivative of a circle

$x^{2} + y^{2} = 25$

was found to be

$\frac{d y}{d x} = - \frac{x}{y}$

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

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (- \frac{x}{y})$

The left side becomes $\frac{d ^{2} y}{d x ^{2}}$ , the second derivative.

$\frac{d ^{2} y}{d x ^{2}} = - (\frac{y ( 1 ) - x ( y ^{'} )}{y ^{2}}) = - \frac{y - x y ^{'}}{y ^{2}}$

Since $y^{'} = - \frac{x}{y}$ , substitute this in:

$\frac{d ^{2} y}{d x ^{2}} = - \frac{y - x ( - \frac{x}{y} )}{y ^{2}} = - \frac{y + \frac{x ^{2}}{y}}{y ^{2}}$

Simplifying,

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

Because the original equation tells us $x^{2} + y^{2} = 25$ , we can also write:

$\frac{d ^{2} y}{d x ^{2}} = - \frac{25}{y ^{3}}$

Example

Evaluate $\frac{d ^{2} y}{d x ^{2}}$ at the point $(1, - 2)$ for

$2 x^{4} - x y = y^{2}$

Solution

Step 1: Find $\frac{d y}{d x}$

(spoiler)

Differentiating implicitly,

$8 x^{3} - (x \cdot \frac{d y}{d x} + y) = 2 y \cdot \frac{d y}{d x}$

Solve for $\frac{d y}{d x}$ by combining like terms:

Step 2: Find $\frac{d ^{2} y}{d x ^{2}}$

(spoiler)

To find the 2nd derivative, implicitly differentiate again while applying the quotient rule. For readability, $\frac{d y}{d x}$ is written as $y^{'}$ .

$\frac{d ^{2} y}{d x ^{2}} = \frac{( x + 2 y ) ( 24 x ^{2} - y ^{'} ) - ( 8 x ^{3} - y ) ( 1 + 2 y ^{'} )}{( x + 2 y ) ^{2}}$

Step 3: Evaluate at the point

(spoiler)

To evaluate at $(1, - 2)$ , we also need the value of $y^{'}$ at the point.

$y^{'} = \frac{8 x ^{3} - y}{x + 2 y} = \frac{8 ( 1 ) ^{3} - ( - 2 )}{1 + 2 ( - 2 )} = \frac{8 + 2}{1 - 4} = - \frac{10}{3}$

Then the value of $\frac{d ^{2} y}{d x ^{2}}$ at $(1, - 2)$ equals

Challenge problem

Consider the curve defined by the equation:

$a x^{2} + b x y + y^{2} = 12$

where $a$ and $b$ are constants.

a) If the tangent line to the curve at the point $(2, 4)$ has a slope of $1,$ find the values of $a$ and $b$ .

b) Evaluate $\frac{d ^{2} y}{d x ^{2}}$ at $(2, 4)$ .

Solutions

a) Find $a$ and $b$ .

(spoiler)

Differentiating implicitly (noting that $a$ and $b$ are constants):

$a (2 x) + b (x \cdot \frac{d y}{d x} + y) + 2 y \cdot \frac{d y}{d x} = 0$

Solving for $\frac{d y}{d x}$ ,

$2 a x + b x \cdot \frac{d y}{d x} + b y + 2 y \cdot \frac{d y}{d x} \frac{d y}{d x} (b x + 2 y) \frac{d y}{d x} = 0 = - 2 a x - b y = \frac{- 2 a x - b y}{b x + 2 y}$

If the tangent line at $(2, 4)$ has a slope of $1$ , then $\frac{d y}{d x} = 1$ when $x = 2$ and $y = 4$ . Substitute these values:

$11 2 b + 8 4 a + 6 b = \frac{- 2 a ( 2 ) - b ( 4 )}{b ( 2 ) + 2 ( 4 )} = \frac{- 4 a - 4 b}{2 b + 8} = - 4 a - 4 b = - 8$

This is one equation relating $a$ and $b$ . To get a second equation, use the equation of the original curve at $(2, 4)$ :

$a x^{2} + b x y + y^{2} a (2)^{2} + b (2) (4) + (4)^{2} 4 a + 8 b = 12 = 12 = - 4$

Solve the system of equations

$4 a + 6 b 4 a + 8 b = - 8 = - 4$

to obtain

$a = - 5 b = 2$

b) Evaluate $\frac{d ^{2} y}{d x ^{2}}$ at $(2, 4)$ .

(spoiler)

Since $a = - 5$ and $b = 2$ ,

$\frac{d y}{d x} = \frac{- 2 a x - b y}{b x + 2 y} = \frac{10 x - 2 y}{2 x - 10 y} = \frac{5 x - y}{x - 5 y}$

Then to find the 2nd derivative, implicitly differentiate:

$\frac{d ^{2} y}{d x ^{2}} = \frac{( x - 5 y ) ( 5 - y ^{'} ) - ( 5 x - y ) ( 1 - 5 y ^{'} )}{( x - 5 y ) ^{2}}$

At $(2, - 4)$ , the 1st derivative is

$y^{'} = \frac{5 ( 2 ) - ( - 4 )}{2 - 5 ( - 4 )} = \frac{10 + 4}{2 + 20} = \frac{14}{22} = \frac{7}{11}$

Then substituting into the 2nd derivative,

Achievable AP Calculus AB

3. Advanced differentiation

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.