Question

Calculate the derivative of the following functions. $$y=(\sec x+\tan x)^{5}$$

Short answer

Question: Find the derivative of the function $$y = (\sec x + \tan x)^{5}$$. Answer: $$\frac{dy}{dx} = 5(\sec x + \tan x)^4(\sec x \tan x + \sec^2 x)$$.

Step-by-step solution

Identify the outer and inner functions

Here, y is a composite function of the form $$y = u^5$$, where $$u = \sec x + \tan x$$. So, the outer function is $$u^5$$ and the inner function is $$\sec x + \tan x$$.

Differentiate the outer function

We need to differentiate the outer function with respect to u. So, $$\frac{dy}{du} = \frac{d(u^5)}{du} = 5u^4$$

Differentiate the inner function

Next, we need to differentiate the inner function with respect to x. So, $$\frac{du}{dx} = \frac{d(\sec x + \tan x)}{dx} = \sec x \tan x + \sec^2 x$$.

Use the Chain Rule to find the derivative of the given function

Now, we'll multiply the derivatives from Step 2 and Step 3 using chain rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ $$\frac{dy}{dx} = (5u^4)(\sec x \tan x + \sec^2 x)$$

Replace u with the inner function

Finally, substitute $$u$$ back in with the inner function $$u = \sec x + \tan x$$: $$\frac{dy}{dx} = 5(\sec x + \tan x)^4(\sec x \tan x + \sec^2 x)$$ The final answer is $$\frac{dy}{dx} = 5(\sec x + \tan x)^4(\sec x \tan x + \sec^2 x)$$.

Key concepts

Derivative

The derivative is a fundamental concept in calculus. It provides a way to measure how a function changes as its input changes. More specifically, it tells us the rate of change or the slope of the function at any particular point. For differentiable functions, the derivative can be thought of as the slope of the tangent line to the graph of the function.

Derivatives are crucial in understanding how quantities vary in real-world contexts, such as speed, growth rates, and much more. To calculate the derivative of a function, we use principles such as the power rule, product rule, and chain rule, depending on the composition of the function.

In our exercise, the function involves trigonometric identities, requiring careful application of derivative rules to each component. Understanding derivatives is not just about calculating them but about interpreting their meaning in terms of how the original function behaves or changes.

Composite Function

A composite function is essentially a function of another function. This means you take the output of one function and use it as the input of another. When you have functions like \( y = (\sec x + \tan x)^5 \), it is a perfect example of a composite function.

You can think of it in terms of layers, where you first calculate the inner function—in this case, \( \sec x + \tan x \)—and then apply another function to the result (the outer function here is raising the expression to the power of 5).

Composite functions require special attention to differentiation because simply applying standard derivative rules isn’t sufficient. You must consider the inner and outer layers separately. This is where the chain rule comes into play, allowing you to differentiate composite functions accurately by breaking them down into simpler parts.

Trigonometric Functions

Trigonometric functions like \( \sec x \) and \( \tan x \) are functions that relate angles of a triangle to the lengths of the sides of the triangle. They are periodic and have particular rules for differentiation. In calculus, they can describe oscillatory behavior, which makes them very useful in modeling cycles and waves, among other applications.

Differentiating these functions requires knowing specific derivatives:

• The derivative of \( \tan x \) is \( \sec^2 x \).
• The derivative of \( \sec x \) is \( \sec x \tan x \).

When these functions are part of a composite function, as in the exercise, their derivatives become part of the inner function's differentiation in the chain rule. Understanding these derivatives helps us work through problems involving trigonometric expressions more systematically. The key is to recall each function's differentiation and apply them effectively during calculations.