Question

1. y = (sec x + tan x)^5

Explanation:

Assuming the problem is to find the derivative of \( y = (\sec x + \tan x)^5 \) (since it's a calculus - related problem, likely finding the derivative):

Step 1: Apply the chain rule

The chain rule states that if we have a function \( y = u^n \), where \( u \) is a function of \( x \) and \( n \) is a constant, then \( \frac{dy}{dx}=n\cdot u^{n - 1}\cdot\frac{du}{dx} \).
Let \( u=\sec x+\tan x \) and \( n = 5 \). Then \( y = u^{5} \), and by the chain rule:
\( \frac{dy}{dx}=5\cdot u^{4}\cdot\frac{du}{dx} \)
Substituting back \( u=\sec x+\tan x \), we get \( \frac{dy}{dx}=5(\sec x + \tan x)^{4}\cdot\frac{d}{dx}(\sec x+\tan x) \)

Step 2: Differentiate \( \sec x+\tan x \)

We know the following derivative rules:

• The derivative of \( \sec x \) with respect to \( x \) is \( \sec x\tan x \), i.e., \( \frac{d}{dx}(\sec x)=\sec x\tan x \)
• The derivative of \( \tan x \) with respect to \( x \) is \( \sec^{2}x \), i.e., \( \frac{d}{dx}(\tan x)=\sec^{2}x \)

Using the sum rule of differentiation (the derivative of a sum of functions is the sum of their derivatives), we have:
\( \frac{d}{dx}(\sec x+\tan x)=\frac{d}{dx}(\sec x)+\frac{d}{dx}(\tan x)=\sec x\tan x+\sec^{2}x \)

Step 3: Simplify the expression

Substitute \( \frac{d}{dx}(\sec x+\tan x)=\sec x\tan x+\sec^{2}x \) back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx}=5(\sec x + \tan x)^{4}(\sec x\tan x+\sec^{2}x) \)
We can factor out \( \sec x \) from the terms \( \sec x\tan x+\sec^{2}x \):
\( \sec x\tan x+\sec^{2}x=\sec x(\tan x + \sec x) \)
So, \( \frac{dy}{dx}=5(\sec x + \tan x)^{4}\cdot\sec x(\sec x+\tan x) \)
Using the rule of exponents \( a^{m}\cdot a^{n}=a^{m + n} \), we get:
\( \frac{dy}{dx}=5\sec x(\sec x + \tan x)^{5} \)

Answer:

The derivative of \( y = (\sec x+\tan x)^{5} \) is \( \boldsymbol{5\sec x(\sec x + \tan x)^{5}} \) (if the problem was to find the derivative. If the problem was different, for example, to expand the function, the answer would be different. But based on the context of calculus and the form of the function, finding the derivative is a common operation)