True or False? Every derivative on this course can be found from the standard rules, without going back to the limit definition.

True. The power, product, quotient and chain rules, together with the standard results, cover every function on the course. The limit definition is what those rules were built from, and it stays the route to take when a derivative has to be justified from first principles.

You are given only a table of values for a function, with no formula for it. What can you still find, and how?

Only an estimate of the derivative at a point, found as the average rate of change between the two nearest values either side of it. Every other method on the course needs a formula for the function to work on.

Which two methods will derive the derivative of an inverse trigonometric function?

The inverse function theorem , and implicit differentiation after rewriting the equation without the inverse. Both finish with a trigonometric identity that removes the inverse function from the answer.

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

Implicit Differentiation (College Board AP® Calculus BC): Flashcards

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FrontImplicit Differentiation
Define an implicit function.

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Define an implicit function.
An implicit function is given by an equation relating $x$ and $y$ that cannot be rearranged to express $y$ as a function of $x$ .
Examples are $3 x^{2} - 7 x y^{2} = 3$ and $x^{2} + y^{2} = 25$ , where the values of $y$ still depend on $x$ without being expressible in terms of it.
Fill in the missing factor in the rule for differentiating a term in $y$ with respect to $x$ :
$\frac{d}{d x} f (y) = f^{'} (y) \cdot______$
The completed rule is $\frac{d}{d x} f (y) = f^{'} (y) \cdot \frac{d y}{d x}$ .
Differentiate the term with respect to $y$ as usual, then multiply by $\frac{d y}{d x}$ , which is the chain rule doing the work.
How do you find $\frac{d y}{d x}$ for a curve given by an implicit equation?
Differentiate every term on both sides with respect to $x$ , treating terms in $y$ by the chain rule.
Then collect the $\frac{d y}{d x}$ terms on one side, factor $\frac{d y}{d x}$ out, and make it the subject.
True or False?
An implicit derivative $\frac{d y}{d x}$ can be evaluated from the $x$ -coordinate of a point alone.
False.
The expression for $\frac{d y}{d x}$ usually involves both $x$ and $y$ , so both coordinates are needed.
Check as well that the point really does lie on the curve, by substituting it into the original equation.
Find $\frac{d y}{d x}$ for the curve $x^{2} + y^{2} = 4 x$ .
It is $\frac{d y}{d x} = \frac{2 - x}{y}$ .
Differentiating gives $2 x + 2 y \frac{d y}{d x} = 4$ , and rearranging gives $\frac{4 - 2 x}{2 y}$ .
What is $\frac{d}{d x} (x y)$ ?
It is $y + x \frac{d y}{d x}$ .
The product rule gives $1 \cdot y + x \cdot \frac{d}{d x} (y)$ , and differentiating $y$ with respect to $x$ contributes the factor $\frac{d y}{d x}$ .
Find $\frac{d y}{d x}$ given that $x^{4} + y^{3} = 12 x y$ .
It is $\frac{d y}{d x} = \frac{12 y - 4 x^{3}}{3 y^{2} - 12 x}$ .
Differentiating gives $4 x^{3} + 3 y^{2} \frac{d y}{d x} = 12 y + 12 x \frac{d y}{d x}$ , and collecting the $\frac{d y}{d x}$ terms produces the result.
How does implicit differentiation give the derivative of $y = \tan^{- 1} x$ ?
Rewrite it as $\tan y = x$ and differentiate both sides, which gives $\sec^{2} y \cdot \frac{d y}{d x} = 1$ .
The identity $\sec^{2} y = \tan^{2} y + 1$ then turns $\frac{1}{\sec^{2} y}$ into $\frac{1}{x^{2} + 1}$ , since $\tan y = x$ .
How does implicit differentiation show that $\frac{d}{d x} (a^{x}) = a^{x} \ln a$ ?
Take logarithms of $y = a^{x}$ to get $\ln y = x \ln a$ , then differentiate both sides with respect to $x$ .
That gives $\frac{1}{y} \cdot \frac{d y}{d x} = \ln a$ , so $\frac{d y}{d x} = y \ln a = a^{x} \ln a$ .
True or False?
Every derivative on this course can be found from the standard rules, without going back to the limit definition.
True.
The power, product, quotient and chain rules, together with the standard results, cover every function on the course.
The limit definition is what those rules were built from, and it stays the route to take when a derivative has to be justified from first principles.
Fill in the three missing rule names:
two functions multiplied together need the $______$ rule, one function divided by another needs the $______$ rule, and one function inside another needs the $______$ rule instead.
The completed statement is: two functions multiplied together need the product rule, one function divided by another needs the quotient rule, and one function inside another needs the chain rule instead.
A single expression will often need two of the three, applied one inside the other.
You are given only a table of values for a function, with no formula for it. What can you still find, and how?
Only an estimate of the derivative at a point, found as the average rate of change between the two nearest values either side of it.
Every other method on the course needs a formula for the function to work on.
Which two methods will derive the derivative of an inverse trigonometric function?
The inverse function theorem, and implicit differentiation after rewriting the equation without the inverse.
Both finish with a trigonometric identity that removes the inverse function from the answer.
Why choose a fresh letter such as $w$ when a second substitution is needed inside a chain rule problem?
Because reusing $u$ or $v$ makes the two layers indistinguishable in the working, and those layers are exactly what the chain rule is keeping track of.
A distinct letter for each layer keeps every derivative attached to the right function.

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