Trigonometric Derivatives

Trigonometric derivatives are mostly formulaic, and require an understanding of trig derivative rules in order to solve differentiation problems involving trig functions. Here are the derivatives of common trigonometric functions:

Example 1: Consider the trigonometric function $f(x) = -\sin(x) - \cos(x)$. Apply trig differentiation rules to compute the derivative.

Solution:

Step 1: Using the rule $\frac{d}{dx}(\sin(x)) = \cos(x)$, the derivative of $-\sin(x)$ is $-\cos(x)$.

Step 2: Using the rule $\frac{d}{dx}(\cos(x)) = -\sin(x)$, the derivative of $-\cos(x)$ is $\sin(x)$.

Now, we have found the derivatives of the individual terms:

$\frac{d}{dx}(-\sin(x)) = -\cos(x)$

$\frac{d}{dx}(-\cos(x)) = \sin(x)$

So, the derivative of $f(x) = -\sin(x) - \cos(x)$ is the sum of the derivatives:

$\boxed{\frac{d}{dx}(-\sin(x) - \cos(x)) = -\cos(x) + \sin(x)}$

Power Rule

The power rule is a fundamental rule that allows us to differentiate functions of the form $f(x) = x^n$, where $n$ is a constant exponent.

The Power Rule states that if $f(x) = x^n$, then:

When we differentiate a term with an exponent $n$, we bring down the exponent as the new coefficient and reduce the exponent by 1. For example, if $g(x) = x^3$, the power rule gives:

Example 2: Consider the function $g(x) = x^{-2}$. Apply the power rule to calculate the derivative.

Solution: Using the power rule:

$g'(x) = -2x^{-2-1} = -2x^{-3} = \frac{-2}{x^3}$

Therefore, the derivative of $g(x) = x^{-2}$ is $\boxed{g'(x) = \dfrac{-2}{x^3}}$.

Product Rule

The product rule allows us to find the derivative of a product of two functions.

The Product Rule states that if $h(x) = f(x) \cdot g(x)$, then:

To find the derivative of a product: take the derivative of the first function times the second, then add the first function times the derivative of the second.

For example, with $f(x) = x^2$ and $g(x) = \sin(x)$:

Example 3: Consider the function $g(x) = (2x + 6)\sin(x)$. Apply the product rule to calculate the derivative.

Solution: Recall the product rule:

$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

Step 1: Identify $f(x)$ and $g(x)$:

$f(x) = 2x + 6, \quad g(x) = \sin(x)$

Step 2: Find the derivatives:

$f'(x) = \frac{d}{dx}(2x + 6) = 2, \quad g'(x) = \frac{d}{dx}(\sin(x)) = \cos(x)$

Step 3: Apply the product rule:

$g'(x) = 2\sin(x) + (2x + 6)\cos(x)$

The derivative of $g(x) = (2x+6)\sin(x)$ is $\boxed{g'(x) = 2\sin(x) + (2x+6)\cos(x)}$.

Quotient Rule

The Quotient Rule states that if $h(x) = \dfrac{f(x)}{g(x)}$, then:

To find the derivative of a quotient: take the derivative of the numerator times the denominator, subtract the numerator times the derivative of the denominator, then divide by the square of the denominator.

For example, with $f(x) = \sin(x)$ and $g(x) = x^2$:

Example 4: Consider the function $h(x) = \dfrac{x^3}{\cos(x)}$. Apply the quotient rule to calculate the derivative.

Solution:

Step 1: Identify $f(x)$ and $g(x)$:

$f(x) = x^3, \quad g(x) = \cos(x)$

Step 2: Find the derivatives:

$(x^3)' = 3x^2, \quad (\cos x)' = -\sin(x)$

Step 3: Apply the quotient rule:

$h'(x) = \frac{3x^2 \cdot \cos(x) - x^3 \cdot (-\sin(x))}{\cos^2(x)}$

Simplifying:

$\boxed{h'(x) = \frac{3x^2\cos(x) + x^3\sin(x)}{\cos^2(x)}}$

Multiple Derivatives

The second derivative, denoted $f''(x)$, represents the rate of change of a function's derivative — in other words, how the slope of a function is changing as we move along the x-axis. It provides information about the curvature and concavity of a function.

Mathematically, the second derivative is the derivative of the first derivative:

This can be extended to a third derivative, and beyond:

Example: For $f(x) = 3x^3 + 2x^2 - 5x + 1$:

Step 1: First derivative:

Step 2: Second derivative:

A positive value of $f''(x)$ indicates the function is concave up $(\cup)$, while a negative value indicates concave down $(\cap)$.

Example 5: Compute the third derivative of $f(x) = \sin(x)$.

Solution:

Step 1: First derivative:

$f'(x) = \frac{d}{dx}(\sin(x)) = \cos(x)$

Step 2: Second derivative:

$f''(x) = \frac{d^2}{dx^2}(\sin(x)) = \frac{d}{dx}(\cos(x)) = -\sin(x)$

Step 3: Third derivative:

$f'''(x) = \frac{d^3}{dx^3}(\sin(x)) = \frac{d}{dx}(-\sin(x)) = \boxed{-\cos(x)}$

Practice Problems