Question

In Exercises \(1-10,\) find \(d y / d x\) . Use your grapher to support your analysis if you are unsure of your answer. $$y=x \sec x$$

Short answer

The derivative, \(d y / d x\), of the function \(y = x \sec x\) is \(d y / d x = \sec x - x \sec x \tan x\)

Step-by-step solution

Express the function

The function to be differentiated is \(y = x \sec x\), where \(\sec x\) is the secant function. One first step is to express that function as \(y = x \cdot (\cos x)^{-1}\) where \(\cos x\) is the cosine function. This is because the secant function is the reciprocal of the cosine function. So, this allows the use of the chain rule when differentiating \((\cos x)^{-1}\).

Apply the product rule

Now apply the product rule to the function \(y = x \cdot (\cos x)^{-1}\). For this, let \(u = x\) and \(v = (\cos x)^{-1}\).

Find u' and v'

Differentiate u and v respectively with respect to x. The derivative of \(x\) (u') is 1. For \((\cos x)^{-1}\) apply the chain rule, The derivative \((\cos x)^{-1}\) with respect to \(x\) is \((v' = -\sin x (\cos x)^{-2})\).

Substitute u', v', u, v into the product rule formula

Using the product rule formula \((uv)' = u'v + uv'\), we substitute the values of \(u, u', v \)and \(v'\) into the formula to get: \((1) \cdot (\cos x)^{-1} + x \cdot (-\sin x (\cos x)^{-2})\).

Finalize

The expression from Step 4 simplifies to \(d y / d x = \sec x - x \sec x \tan x\).

Key concepts

Product Rule in Differential Calculus

In differential calculus, the product rule is a technique used for finding the derivative of a function that is the product of two functions. It is a foundational tool that allows us to tackle complex functions composed of simpler parts.

For example, suppose we want to differentiate a function in the form of \( f(x) = g(x)h(x) \), where both \( g(x) \) and \( h(x) \) are differentiable functions. Using the product rule, the derivative of \( f(x) \) is given by: \[ f'(x) = g'(x)h(x) + g(x)h'(x) \]

This formula states that the derivative of the product of two functions is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function. In the context of the given exercise, where we have \( y = x \sec(x) \), identifying \( u = x \) and \( v = \sec(x) \), and knowing their derivatives are \( u' = 1 \) and \( v' = \sec(x)\tan(x) \) respectively, we use the product rule to arrive at the solution for \( \frac{dy}{dx} \).

It's crucial to apply the product rule correctly, taking care to differentiate each function correctly before substituting them into the formula. This ensures accuracy in finding derivatives of products of functions.

Derivative of the Secant Function

The secant function, represented as \( \text{sec}(x) \), is one of the six fundamental trigonometric functions. In calculus, understanding how to differentiate trigonometric functions is essential. When we differentiate the secant function, we apply the rules of calculus to trigonometric identities.

The derivative of \( \sec(x) \) can be derived from its relationship to the cosine function, as \( \sec(x) = \frac{1}{\cos(x)} \). Using this relationship and the chain rule (which will be discussed in the next section), the derivative of \( \sec(x) \) is given by: \[ \frac{d}{dx}\sec(x) = \sec(x)\tan(x) \]

The process to find this derivative involves differentiating the inverse cosine within the context of the function composition, a technique made accessible by the chain rule. In our exercise, this knowledge allows us to differentiate \( x \sec(x) \), as the secant function's derivative informs the second part of the product rule application.

Chain Rule and Its Application

The chain rule is a principle in calculus used to find the derivative of a composition of functions. It is instrumental when dealing with functions within functions, a common occurrence in calculus problems.

Expressed formally, if we have a composite function \( f(g(x)) \), and wish to find \( f'(g(x)) \), we would compute the following: \[ f'(g(x)) = f'(g(x)) \cdot g'(x) \]

In other words, the derivative of a composite function is the derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function.

Back to our problem at hand: the exercise required applying the chain rule to differentiate \( (\cos(x))^{-1} \). This entails treating \( (\cos(x))^{-1} \) as the outer function, and \( \cos(x) \) as the inner function. The derivative of the outer function \( u^{-1} \) is \( -u^{-2} \), and when applied to \( \cos(x) \), we multiply it by the derivative of \( \cos(x) \), which is \( -\sin(x) \), leading to the final derivative of the inner function \( -\sin(x) \) times the square of the secant function. This illustrates the power of the chain rule: it enables us to differentiate complex functions by breaking them down into their constituent parts.