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

Higher-order implicit derivatives: Find the second derivative of an implicitly defined relation by substituting the first derivative.

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.

In the previous section, implicitly differentiating the circle $x^{2} + y^{2} = 25$ gave the 1st derivative:

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

To find the 2nd 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 $x$ .

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

The expression still contains the term $\frac{d y}{d x}$ . Replace it using the original first derivative equation ( $\frac{d y}{d x} = - \frac{x}{y}$ ):

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

To simplify the complex fraction, multiply both the entire numerator and the denominator by $y$ :

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

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

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

Example

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

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

Solution

(spoiler)

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

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:

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

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

To find the 2nd derivative, implicitly differentiate again while applying the quotient rule.

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

3. Evaluate at the point:

To evaluate at $(1, - 2)$ , we also need the value of $\frac{d y}{d x}$ at the point.

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

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

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

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

$a b = - 5 = 2$

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:

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

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

$\frac{d y}{d x} = \frac{5 ( 2 ) - ( - 4 )}{2 - 5 ( - 4 )} = \frac{7}{11}$

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

Higher-order implicit derivatives: Find the second derivative of an implicitly defined relation by substituting the first derivative.