Define the second derivative of a function.

The second derivative is the derivative of the derivative, found by differentiating the function and then differentiating the result. It measures the rate of change of the rate of change, so it says how fast the first derivative is itself rising or falling.

True or False? Any function that can be differentiated once can be differentiated a second time.

False. The second derivative exists only where the first derivative is itself differentiable, and a derivative can perfectly well have a corner or a break in it. Each further order needs the one before it to be differentiable all over again.

A function gives the displacement of an object. What do its first, second and third derivatives describe?

The first derivative gives the velocity and the second gives the acceleration . The third gives the rate of change of acceleration, which is sometimes called jerk.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 3: Differentiation of Composite, Implicit & Inverse FunctionsDifferentiation of Composite & Inverse Functions

Differentiation of Composite & Inverse Functions (College Board AP® Calculus BC): Flashcards

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FrontThe Chain Rule
If $y = f (u)$ and $u = g (x)$ , fill in the two missing derivatives in the chain rule:
$\frac{d y}{d x} =______\cdot______$

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Cards in this collection (28)

If $y = f (u)$ and $u = g (x)$ , fill in the two missing derivatives in the chain rule:
$\frac{d y}{d x} =______\cdot______$
The completed rule is $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ .
These are derivatives rather than fractions, but the $d u$ terms appear to cancel, which is a useful check that you have written the rule the right way round.
What is the chain rule in function notation for $h (x) = f (g (x))$ ?
It is $h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$ .
Differentiate the outer function, leave the inner function untouched inside it, then multiply by the derivative of that inner function.
True or False?
The derivative of $\sin (x^{2})$ is $\cos (x^{2})$ .
False.
The derivative of the inner function has been left out.
The chain rule gives $2 x \cos (x^{2})$ , since the inner function $x^{2}$ has derivative $2 x$ .
Differentiate $y = {(2 x^{3} + 4 x)}^{6}$ using the chain rule.
The derivative is $\frac{d y}{d x} = 6 {(2 x^{3} + 4 x)}^{5} (6 x^{2} + 4)$ .
With $u = 2 x^{3} + 4 x$ the function is $y = u^{6}$ , so $\frac{d y}{d u} = 6 u^{5}$ and $\frac{d u}{d x} = 6 x^{2} + 4$ .
A table gives $g (2) = 5$ , $g^{'} (2) = 4$ and $f^{'} (5) = - 3$ . Find $h^{'} (2)$ where $h (x) = f (g (x))$ .
It is $- 12$ .
The chain rule gives $h^{'} (2) = f^{'} (g (2)) \cdot g^{'} (2) = f^{'} (5) \cdot 4$ .
How do you differentiate a function nested three deep, such as $\sin (e^{3 x^{2} + 2})$ ?
Apply the chain rule twice, which amounts to $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x}$ with two substitutions.
Use a fresh letter for each substitution, so that the layers do not get confused with one another.
Differentiate $g (x) = e^{\tan x}$ .
The derivative is $g^{'} (x) = e^{\tan x} \sec^{2} x$ .
The outer function is the exponential, which differentiates to itself, and the inner function $\tan x$ contributes the factor $\sec^{2} x$ .
Differentiating $f (x) = \cos x \cdot e^{4 x^{3} + 3 x^{2} + 2}$ needs two rules. Which, and in what order?
Start with the product rule, taking $u = \cos x$ and $v = e^{4 x^{3} + 3 x^{2} + 2}$ .
The chain rule is then needed inside it, to differentiate $v$ , which gives $v^{'} = (12 x^{2} + 6 x) e^{4 x^{3} + 3 x^{2} + 2}$ .
Fill in the missing derivative and the missing condition:
$\frac{1}{(\frac{d y}{d x})} =______provided that \frac{d y}{d x} \neq______$
The completed statement is $\frac{1}{(\frac{d y}{d x})} = \frac{d x}{d y}$ provided that $\frac{d y}{d x} \neq 0$ .
Derivatives are not fractions, but they do behave like fractions when reciprocals are taken.
State the inverse function theorem for the derivative of $f^{- 1}$ at $a$ .
It is ${(f^{- 1})}^{'} (a) = \frac{1}{f^{'} (f^{- 1} (a))}$ , provided that $f^{'} (f^{- 1} (a)) \neq 0$ .
Writing $g$ for $f^{- 1}$ turns it into the shorter $g^{'} (a) = \frac{1}{f^{'} (g (a))}$ .
True or False?
The form $\frac{d y}{d x} = \frac{1}{(\frac{d x}{d y})}$ and the form ${(f^{- 1})}^{'} (a) = \frac{1}{f^{'} (f^{- 1} (a))}$ state the same theorem.
True.
They differ only in notation and in what each is convenient for.
The first gives an expression for the derivative of the inverse, though in terms of $y$ , while the second gives its value at a point.
What must be true of $f$ before the inverse function theorem can be used?
It must be differentiable, and it must be one-to-one so that an inverse exists at all.
The derivative of $f$ at the matching point must also be non-zero, since that value ends up in a denominator.
Why is the slope of $f^{- 1}$ the reciprocal of the slope of $f$ at the matching point?
Because the graph of $f^{- 1}$ is the reflection of the graph of $f$ in the line $y = x$ , which swaps the horizontal and vertical changes over.
A slope of $\frac{q}{p}$ on one graph therefore appears as $\frac{p}{q}$ on the other.
How does the chain rule give the inverse function theorem?
Start from $f (g (x)) = x$ , which holds because $g$ is the inverse of $f$ .
Differentiating both sides gives $f^{'} (g (x)) \cdot g^{'} (x) = 1$ , and rearranging gives $g^{'} (x) = \frac{1}{f^{'} (g (x))}$ .
Let $f (x) = {(3 x + 4)}^{4}$ with inverse $g$ , and let $f (0) = 256$ . Find $g^{'} (256)$ .
It is $\frac{1}{768}$ .
Since $f (0) = 256$ we have $g (256) = 0$ , and $f^{'} (x) = 12 {(3 x + 4)}^{3}$ gives $f^{'} (0) = 768$ .
Fill in the three missing derivatives:
$\frac{d}{d x} (\sin^{- 1} x) =______$
$\frac{d}{d x} (\cos^{- 1} x) =______$
$\frac{d}{d x} (\tan^{- 1} x) =______$
The completed results are $\frac{1}{\sqrt{1 - x^{2}}}$ , then $- \frac{1}{\sqrt{1 - x^{2}}}$ , then $\frac{1}{1 + x^{2}}$ .
Each inverse co-function derivative is its partner's derivative with a minus sign in front, and that pattern holds for all three pairs.
How does the inverse function theorem give the derivative of $\sin^{- 1} x$ ?
With $f (x) = \sin x$ the theorem gives $\frac{1}{\cos (\sin^{- 1} x)}$ .
The identity $\cos θ = \sqrt{1 - \sin^{2} θ}$ then turns the denominator into $\sqrt{1 - x^{2}}$ , because $\sin (\sin^{- 1} x) = x$ .
True or False?
$\sin^{- 1} x$ and its derivative are defined on exactly the same set of values.
False.
The function is defined for $- 1 \leq x \leq 1$ , but its derivative $\frac{1}{\sqrt{1 - x^{2}}}$ is defined only for $- 1 < x < 1$ .
At the two endpoints the denominator is zero and the derivative becomes unbounded.
In deriving the derivative of $\sin^{- 1} x$ , why is the negative square root discarded?
Because $\sin^{- 1} x$ takes values in $[- \frac{π}{2}, \frac{π}{2}]$ , and $\cos$ is never negative on that interval.
The graph of $y = \sin^{- 1} x$ also has positive slope throughout, which the negative root would contradict.
Fill in the two missing derivatives:
$\frac{d}{d x} (\sec^{- 1} x) =______$
$\frac{d}{d x} (\cot^{- 1} x) =______$
The completed results are $\frac{1}{| x | \sqrt{x^{2} - 1}}$ for the inverse secant and $- \frac{1}{1 + x^{2}}$ for the inverse cotangent.
The first is defined only where $x < - 1$ or $x > 1$ , while the inverse secant itself is also defined at $x = \pm 1$ .
Differentiate $f (x) = \arcsin (2 x^{3} + e^{2 x})$ .
The derivative is $f^{'} (x) = \frac{6 x^{2} + 2 e^{2 x}}{\sqrt{1 - {(2 x^{3} + e^{2 x})}^{2}}}$ .
The chain rule takes the standard inverse sine derivative with the inner function in place of $x$ , and multiplies it by that inner function's own derivative.
Why is $\frac{d}{d x} (\tan^{- 1} x)$ defined for every real $x$ , unlike the derivative of $\sin^{- 1} x$ ?
Because its denominator $1 + x^{2}$ is never zero, whereas $\sqrt{1 - x^{2}}$ vanishes at $x = \pm 1$ .
The inverse tangent is itself defined for every real number, so there are no endpoints for its derivative to fail at.
Define the second derivative of a function.
The second derivative is the derivative of the derivative, found by differentiating the function and then differentiating the result.
It measures the rate of change of the rate of change, so it says how fast the first derivative is itself rising or falling.
Find the second derivative of $f (x) = 3 x^{3} - 2 x^{2} - 3 x + 2$ .
It is $f^{''} (x) = 18 x - 4$ .
The first derivative is $9 x^{2} - 4 x - 3$ , and differentiating that again gives the answer.
True or False?
Any function that can be differentiated once can be differentiated a second time.
False.
The second derivative exists only where the first derivative is itself differentiable, and a derivative can perfectly well have a corner or a break in it.
Each further order needs the one before it to be differentiable all over again.
For $y = f (x)$ , fill in the two missing notations for the second derivative:
$f^{''} (x)$ and $______$ and also $______$ as a third form.
The completed list is $f^{''} (x)$ and $\frac{d^{2} y}{d x^{2}}$ and also $y^{''}$ as a third form.
Note where the exponents sit in the Leibniz notation: above the $d$ on the top, and after the $x$ underneath.
How are the third and fourth derivatives written in prime notation?
They are written $f^{(3)} (x)$ and $f^{(4)} (x)$ , with the order shown in brackets.
You may occasionally meet $f^{'''} (x)$ for the third, but beyond that the primes are almost never used.
A function gives the displacement of an object. What do its first, second and third derivatives describe?
The first derivative gives the velocity and the second gives the acceleration.
The third gives the rate of change of acceleration, which is sometimes called jerk.

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