$\tan x = \frac{\sin x}{\cos x}$

$\cot x = \frac{\cos x}{\sin x} = \frac{1}{\tan x}$

$\sec x = \frac{1}{\cos x}$

$\csc x = \frac{1}{\sin x}$

The instantaneous rate of change of a function.

A formula for finding the derivative of a composite function.

A function that is composed of another function. For example, $f(g(x))$.

A formula used to differentiate products of two or more functions.

A method of finding the derivative of a function that is the ratio of two other functions.

Functions that relate the angles of a triangle to the lengths of its sides.

1. Find the derivative of $2\tan(x)$ which is $2\sec^2(x)$. 2. Find the derivative of $\sec(x)$ which is $\sec(x)\tan(x)$. 3. Add the derivatives together: $2\sec^2(x) + \sec(x)\tan(x)$.

1. Apply the quotient rule: $\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$. 2. Identify $u(x) = \cot(x)$ and $v(x) = \csc(x)$. 3. Find $u'(x) = -\csc^2(x)$ and $v'(x) = -\csc(x)\cot(x)$. 4. Substitute into the quotient rule: $\frac{-\csc^2(x) \cdot \csc(x) - \cot(x) \cdot [-\csc(x)\cot(x)]}{\csc^2(x)}$. 5. Simplify: $\frac{-\csc^3(x) + \csc(x)\cot^2(x)}{\csc^2(x)}$. 6. Further simplify to $-\csc(x) + \frac{\cot^2(x)}{\csc(x)} = -\csc(x) + \frac{\cos^2(x)}{\sin(x)} \cdot \frac{\sin(x)}{1} = -\csc(x) + \frac{\cos^2(x)}{\sin(x)} = -\csc^2(x)$

1. Apply the chain rule. 2. Let $u = \tan(6x)$, so $g(x) = u^2$. 3. Find $\frac{du}{dx} = 6\sec^2(6x)$. 4. Find $\frac{dg}{du} = 2u = 2\tan(6x)$. 5. Multiply: $\frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = 2\tan(6x) \cdot 6\sec^2(6x) = 12\tan(6x)\sec^2(6x)$.

1. Use the constant multiple rule: $\frac{d}{dx} [cf(x)] = c \frac{d}{dx} f(x)$. 2. Find the derivative of $\cot(x)$, which is $-\csc^2(x)$. 3. Multiply by the constant: $5(-\csc^2(x)) = -5\csc^2(x)$.

1. Find the derivative of the function. 2. Evaluate the derivative at the given x-value to find the slope of the tangent line. 3. Find the y-value of the original function at the given x-value. 4. Use the point-slope form of a line to write the equation of the tangent line.

1. Find the derivative of the function. 2. Set the derivative equal to zero and solve for x. 3. Find where the derivative is undefined.

1. Find the critical points of the function. 2. Evaluate the function at the critical points and endpoints of the interval. 3. The largest value is the maximum, and the smallest value is the minimum.

1. Identify the numerator $u(x)$ and the denominator $v(x)$. 2. Find the derivatives $u'(x)$ and $v'(x)$. 3. Apply the formula: $\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

1. Identify the two functions $u(x)$ and $v(x)$. 2. Find the derivatives $u'(x)$ and $v'(x)$. 3. Apply the formula: $\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.

1. Identify the outer function $f(u)$ and the inner function $g(x)$. 2. Find the derivatives $f'(u)$ and $g'(x)$. 3. Apply the formula: $\frac{d}{dx} f(g(x)) = f'(g(x)) cdot g'(x)$.

$(\tan x)' = \sec^2 x$

$(\cot x)' = -\csc^2 x$

$(\sec x)' = \sec x \tan x$

$(\csc x)' = -\csc x \cot x$

$\frac{d}{dx} f(g(x)) = f'(g(x)) cdot g'(x)$

$\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

$\frac{d}{dx} x^n = nx^{n-1}$

$\frac{d}{dx} c = 0$

$\frac{d}{dx} cf(x) = c \frac{d}{dx} f(x)$