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

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

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

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

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

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

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

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

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

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

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

The derivatives of co-functions (cot, csc) are negative because their original functions are decreasing over their primary intervals.

The chain rule is crucial for finding derivatives of composite functions, where one function is nested inside another. It ensures that we account for the rate of change of both the outer and inner functions.

Simplifying trig expressions before differentiating can make the process easier by reducing complexity. Use trig identities to simplify before applying derivative rules.

The unit circle provides a visual representation of trigonometric functions, helping to understand their behavior and derivatives. It shows how sine, cosine, and their related functions change as the angle varies.

The derivative of $\tan x$ is $\sec^2 x$. This means that the rate of change of the tangent function is related to the square of the secant function.

The derivative of $\cot x$ is $-\csc^2 x$. This means that the rate of change of the cotangent function is related to the negative of the square of the cosecant function.

Using radians ensures that the standard derivative formulas for trigonometric functions are valid. If degrees are used, a conversion factor is needed, making the calculations more complex.

The sign of the derivative indicates whether a function is increasing (positive derivative) or decreasing (negative derivative).

Derivatives are used to find critical points (where the derivative is zero or undefined) of a function, which can then be used to determine the maximum or minimum values of the function.

The derivative of a function gives the slope of the tangent line to the function at any given point. It represents the instantaneous rate of change of the function.