Question

Find the derivative of the following functions. $$y=\sec x \tan x$$

Short answer

Answer: The derivative of the function $$y = \sec x \tan x$$ is $$y' = \sec x \tan^2 x + \sec^3 x$$.

Step-by-step solution

Apply the product rule

The product rule states that if we have a function $$y = u(x) \cdot v(x)$$, the derivative $$y'$$ is given by $$y'= u'(x) \cdot v(x) + u(x) \cdot v'(x)$$. In this case, $$u(x) = \sec x$$ and $$v(x) = \tan x$$, so we need to find the derivatives of these two functions.

Find the derivative of secant function

The derivative of the secant function is given by the formula: $$\frac{d}{dx} (\sec x) = \sec x \tan x$$ In our problem, $$u'(x) = \frac{d}{dx} (\sec x) = \sec x \tan x$$.

Find the derivative of tangent function

The derivative of the tangent function is given by the formula: $$\frac{d}{dx} (\tan x) = \sec^2 x$$ In our problem, $$v'(x) = \frac{d}{dx} (\tan x) = \sec^2 x$$.

Apply the product rule

Now that we have the derivatives of $$u(x)$$ and $$v(x)$$, we can apply the product rule to find the derivative of $$y$$: $$y'= u'(x) \cdot v(x) + u(x) \cdot v'(x)$$ Substitute the derivatives and functions we found in steps 2 and 3: $$y'= (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x)$$

Simplify the expression

We will now simplify the expression for $$y'$$: $$y'= \sec x \tan^2 x + \sec^3 x$$ This is the derivative of the given function $$y = \sec x \tan x$$.

Key concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions that are products of two or more component functions. When dealing with functions of the form \( y = u(x) \cdot v(x) \), the product rule provides a reliable method to find their derivative. The formula for the product rule is as follows:

• \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

To apply the product rule, we need to determine the derivatives of each part of the function separately. In scenarios where one function is multiplied by another, understanding how each part behaves independently helps predict how the entire product changes. This approach breaks a complex task into simpler parts, making calculus problems more manageable.
This rule is particularly handy when dealing with trigonometric functions, such as secant and tangent, which often occur in pairs or products in calculus problems.

Secant Function

The secant function, \( \sec x \), is a trigonometric function related to the cosine function. It is defined as the reciprocal of the cosine function, which means \( \sec x = \frac{1}{\cos x} \). The derivative of the secant function is crucial when applying the product rule in trigonometric identities.

• The formula for the derivative of the secant function is:
  \( \frac{d}{dx} (\sec x) = \sec x \tan x \)

This derivative is useful as it often combines with or simplifies other derivatives in complex expressions. Knowing how to differentiate secant introduces efficiency in simplifying trigonometric derivatives and solving calculus problems that involve these functions.

Tangent Function

The tangent function, denoted as \( \tan x \), measures the ratio of the sine to the cosine of an angle in a triangular context. In calculus, particularly for derivatives, \( \tan x \) also features prominently alongside other trigonometric functions.

• The derivative of the tangent function is:
  \( \frac{d}{dx} (\tan x) = \sec^2 x \)

Understanding this derivative helps simplify expressions in calculus—especially when differentiating functions involving the tangent function. It also provides an essential building block in computing derivatives of trigonometric identities when applying differentiation rules like the product rule.

Derivative Simplification

Once the derivatives of individual components of a function have been found using the product rule, the next step is simplification. This process involves combining and reducing the terms in the expression to achieve a cleaner, more interpretable result.
After applying the product rule to \( y = \sec x \tan x \) and finding the derivatives \( u'(x) = \sec x \tan x \) and \( v'(x) = \sec^2 x \), you substitute these back into the derivative formula:

• \( y' = (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x) \)

Finally, simplify by collecting like terms or using algebraic identities to reduce the expression. In this exercise, the simplified derivative becomes \( y' = \sec x \tan^2 x + \sec^3 x \). Simplification in derivatives not only makes the function easier to understand but also prepares it for further analysis or integration.