Unit 3: Differentiation: Composite, Implicit and Inverse Functions

15 terms. Say what each one means, then open the card to check.

• Composite function

  A function made by putting one function inside another: $f(g(x))$ applies $g$ first, then $f$. The outer function is $f$ and the inner function is $g$.

  Example. $\sqrt{3x + 1}$ is $f(g(x))$ with outer $f(u) = \sqrt{u}$ and inner $g(x) = 3x + 1$.

• Chain rule

  $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$: the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function.

  Example. $\frac{d}{dx}(5x - 2)^4 = 4(5x - 2)^3 \cdot 5 = 20(5x - 2)^3$.

• Chain rule in Leibniz form

  If $y$ depends on $u$ and $u$ depends on $x$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. Rates multiply along the chain.

  Example. If $\frac{dy}{du} = 3$ and $\frac{du}{dx} = 4$ at matching values of $u$ and $x$, then $\frac{dy}{dx} = 12$ there.

• Exponential chain rule

  $\frac{d}{dx}e^{g(x)} = e^{g(x)}g'(x)$, and for a base $b > 0$, $\frac{d}{dx}b^x = b^x\ln b$.

  Example. $\frac{d}{dx}e^{\sin x} = \cos x \cdot e^{\sin x}$ and $\frac{d}{dx}10^x = 10^x\ln 10$.

• Logarithm chain rule

  $\frac{d}{dx}\ln g(x) = \frac{g'(x)}{g(x)}$ wherever $g(x) > 0$, and $\frac{d}{dx}\log_b x = \frac{1}{x\ln b}$ for a base $b > 0$, $b \ne 1$.

  Example. $\frac{d}{dx}\ln(x^2 + 4) = \frac{2x}{x^2 + 4}$.

• Implicit differentiation

  Differentiating both sides of an equation in $x$ and $y$ with respect to $x$, treating $y$ as a function of $x$, then solving for $\frac{dy}{dx}$.

  Example. From $xy = 6$, the product rule gives $y + x\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{y}{x}$.

• The $\frac{dy}{dx}$ factor

  Every time you differentiate a term with $y$ in it, the chain rule attaches a factor of $\frac{dy}{dx}$, because $y$ depends on $x$.

  Example. $\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$, while $\frac{d}{dx}(x^3) = 3x^2$.

• Horizontal and vertical tangents

  On a curve with $\frac{dy}{dx} = \frac{N}{D}$, a horizontal tangent occurs where $N = 0$ and $D \ne 0$, and a vertical tangent where $D = 0$ and $N \ne 0$.

  Example. On $9x^2 + y^2 = 9$, $\frac{dy}{dx} = -\frac{9x}{y}$, so the tangent is horizontal at $(0, 3)$ and vertical at $(1, 0)$.

• Inverse function

  For a one-to-one function $f$, the function $g = f^{-1}$ that undoes it: $g(f(x)) = x$. If $f(a) = b$, then $g(b) = a$, and the graphs are reflections over $y = x$.

  Example. If $f(3) = 10$, then $f^{-1}(10) = 3$.

• Derivative of an inverse

  If $g = f^{-1}$ and $f(a) = b$ with $f'(a) \ne 0$, then $g'(b) = \frac{1}{f'(a)}$: the reciprocal of the slope of $f$ at the matching point.

  Example. If $f(2) = 9$ and $f'(2) = 7$, then $(f^{-1})'(9) = \frac{1}{7}$.

• Derivative of arcsine

  $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$. Arccosine has the same derivative with a minus sign.

  Example. At $x = 0$, the slope of $y = \arcsin x$ is $\frac{1}{\sqrt{1}} = 1$.

• Derivative of arctangent

  $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ for every real $x$.

  Example. At $x = 2$, the slope of $y = \arctan x$ is $\frac{1}{1 + 4} = \frac{1}{5}$.

• Second derivative

  The derivative of the derivative, written $f''(x)$ or $\frac{d^2y}{dx^2}$. It is the rate at which the slope changes.

  Example. For $f(x) = x^4$, $f'(x) = 4x^3$ and $f''(x) = 12x^2$, so $f''(1) = 12$.

• Higher-order derivatives

  Keep differentiating: $f'''$ is the third derivative and $f^{(n)}$ is the $n$th. Each one is the rate of change of the one before.

  Example. For $f(x) = e^{2x}$, each derivative doubles: $f^{(n)}(x) = 2^n e^{2x}$, so $f'''(0) = 8$.

• Second derivative found implicitly

  Differentiate $\frac{dy}{dx}$ again with respect to $x$, then replace every $\frac{dy}{dx}$ that appears with its formula.

  Example. On $xy = 1$, $\frac{dy}{dx} = -\frac{y}{x}$, and differentiating again gives $\frac{d^2y}{dx^2} = \frac{2y}{x^2}$, which is 2 at $(1, 1)$.