Unit 3

Differentiation: Composite, Implicit, and Inverse Functions: the one-page sheet

3.1 The Chain Rule

For a composite function f(g(x)) , differentiate the outside function at the unchanged inside, then multiply by the derivative of the inside: d/dx f(g(x)) = f′(g(x))⋅g′(x) . Name the layers before you start, and never drop the inside derivative.

Spot the layers first: name the outside function and the inside function before you differentiate.
Differentiate the outside function and leave the inside alone: $f'(g(x))$ , not $f'(x)$ and not $f'(g'(x))$ .
Then multiply by the derivative of the inside, $g'(x)$ . Forgetting this factor is the most common lost point.
From a table, read $f'$ at the inside value $g(a)$ , then multiply by $g'(a)$ .

x feeds the inside function g , and g(x) feeds the outside function f and f changes f′(g(x)) times as much as its input g does h′(x) = f′(g(x))⋅g′(x) dy/dx = dy/du ⋅du/dx

chain rule: The rule for differentiating a composite function: $\frac{d}{dx} f(g(x))$ $= f'(g(x)) \cdot g'(x)$ . The derivative of the outside function $f$ , evaluated at the inside function $g(x)$ , times the derivative of the inside function.

3.2 Implicit Differentiation

To differentiate an equation in x and y , apply d/dx to both sides, treating y as a function of x : each y -term gets a factor dy/dx , and products need the product rule. Collect the dy/dx terms, factor, and solve. Substitute a point of the curve to get a slope.

Every term with $y$ gets a factor $\frac{dy}{dx}$ : $\frac{d}{dx}y^2 = 2y\frac{dy}{dx}$ , $\frac{d}{dx}\sin y = \cos y\frac{dy}{dx}$ .
A product of $x$ and $y$ needs the product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ .
Write the whole equation after differentiating, with $= 0$ or the right side, before you solve for $\frac{dy}{dx}$ .
The answer usually holds both $x$ and $y$ ; a slope needs both coordinates of a point on the curve.

near a point, y still acts like a function of x each term with y gets a factor dy/dx by the chain rule; products like xy need the product rule dy/dx equals a fraction that may hold both x and y the result is the slope of the tangent line there

implicit differentiation: Finding $\frac{dy}{dx}$ from an equation in $x$ and $y$ without solving for $y$ : differentiate both sides with respect to $x$ , using the chain rule on every term that holds $y$ , then solve for $\frac{dy}{dx}$ .

3.3 Derivatives of Inverse Functions

The graph of an inverse is a reflection in y = x , which swaps rise and run, so slopes become reciprocals at matching points: (f⁻¹)′(a) = 1/(f′(f⁻¹(a))) . Find f⁻¹(a) first, then take the reciprocal of f′ at that input.

The rule: $g'(x) = \frac{1}{f'(g(x))}$ when $g$ is the inverse of $f$ and $f'(g(x)) \ne 0$ .
Two steps: first find $g(x)$ , the input of $f$ that gives $x$ ; then take the reciprocal of $f'$ there.
From a table, find the given number in the $f(x)$ row, read its $x$ , then read $f'$ in that column.
It comes from the chain rule: differentiate $f(g(x)) = x$ to get $f'(g(x)) \cdot g'(x) = 1$ .

the point (a, b) on f becomes (b, a) on g a tangent line of slope m becomes one of slope 1/m g′(b) = 1/(f′(g(b))) the tangent line of g at b is vertical and g′(b) does not exist

derivative of an inverse function: The derivative of the inverse of $f$ : $\left(f^{-1}\right)'(a)$ $= \frac{1}{f'\left(f^{-1}(a)\right)}$ , provided $f'\left(f^{-1}(a)\right) \ne 0$ . The slope of the inverse at a point is the reciprocal of the slope of $f$ at the reflected point.

3.4 Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions are algebraic: d/dx arcsin x = 1/(√(1 − x²)) , d/dx arccos x = −1/(√(1 − x²)) and d/dx arctan x = 1/(1 + x²) . With an inside function, square the whole inside and multiply by its derivative.

$\frac{d}{dx}\arcsin x$ $= \frac{1}{\sqrt{1 - x^2}}$ and $\frac{d}{dx}\arccos x$ $= -\frac{1}{\sqrt{1 - x^2}}$ , for $-1 < x < 1$ .
$\frac{d}{dx}\arctan x$ $= \frac{1}{1 + x^2}$ for every $x$ .
With an inside function, square the whole inside in the formula and multiply by its derivative.
The results come from implicit differentiation or from the inverse rule $\frac{1}{f'(g(x))}$ .

differentiate implicitly: cos y⋅dy/dx = 1 d/dx arcsin x = 1/(√(1 − x²)) d/dx arctan x = 1/(1 + x²) multiply by u′ : d/dx arctan u = u′/(1 + u²)

derivative of arcsine: The derivatives of the inverse trigonometric functions: $\frac{d}{dx}\arcsin x$ $= \frac{1}{\sqrt{1 - x^2}}$ and $\frac{d}{dx}\arccos x$ $= -\frac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$ , and $\frac{d}{dx}\arctan x$ $= \frac{1}{1 + x^2}$ .

3.5 Choosing a Differentiation Rule

Choose the first rule from the last operation: a sum term by term, a product by the product rule, a quotient by the quotient rule (or a rewrite), a function of a function by the chain rule. Then ask the same question of each piece, one rule per layer.

The last operation decides the first rule: $x^2\sin(3x)$ is a product; $\sin^2(x^3)$ is a composition (a square).
Inside a product or quotient, any factor may still need the chain rule.
Equations mixing $x$ and $y$ call for implicit differentiation; an inverse calls for $\frac{1}{f'(g(x))}$ .
Rewrite before differentiating when it saves a rule, but do not simplify more than the question asks.

that last operation names the first rule: sum, product, quotient or composition it leaves smaller pieces, each needing its own derivative one rule per layer, until only basic derivatives remain 5/x³ = 5x⁻³ needs only the power rule

differentiation strategy: Choosing which differentiation rule to apply first by finding the last operation in the expression (sum, product, quotient or composition), then repeating the choice on each piece; rewriting first when it removes a rule.

3.6 Second and Higher Derivatives

The second derivative f″ is the derivative of f′ : it tells how fast the rate of change is changing. Find it by differentiating again, with the product and chain rules as needed; from a graph of f′ , read it as a slope. For implicit curves, substitute dy/dx after the second differentiation.

Notation: $f''(x)$ , $y''$ , $\frac{d^2y}{dx^2}$ ; higher ones are $f'''(x)$ and $f^{(n)}(x)$ .
Read $f''$ from a graph of $f'$ as its slope, not its height.
If $f'$ is a product or composition, differentiating again needs the product or chain rule again.
Implicit second derivative: differentiate $\frac{dy}{dx}$ (quotient rule, $y$ still depends on $x$ ), then substitute $\frac{dy}{dx}$ .

it can be differentiated: f″ = (f′)′ f″ > 0 where f′ rises, f″ < 0 where f′ falls f‴ , f^((4)) , ..., f^((n)) ; for a polynomial of degree n , f^((n + 1)) = 0 then substitute the expression for dy/dx wherever it appears

second derivative: The derivative of the derivative: $f''(x) = \frac{d}{dx}f'(x)$ , also written $\frac{d^2y}{dx^2}$ . It measures how fast $f'$ changes. Higher-order derivatives continue: $f'''$ , $f^{(4)}$ , ..., $f^{(n)}$ .