Problem 8 - Thomas Calculus

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\). $$ y=-\sec u, \quad u=x^{2}+7 x $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = -\sec(x^2 + 7x) \tan(x^2 + 7x) (2x + 7)\).

Step 1: Find Derivative of y with Respect to u

Given that \(y = -\sec u\), we need to find \(\frac{dy}{du}\). Recall the derivative of \(\sec u\) is \(\sec u \tan u\). Thus, \(\frac{dy}{du} = -\sec u\tan u\).

Step 2: Find Derivative of u with Respect to x

We are given \(u = x^2 + 7x\). To find \(\frac{du}{dx}\), we differentiate \(u\) with respect to \(x\). Applying the power rule, \(\frac{du}{dx} = 2x + 7\).

Step 3: Apply the Chain Rule to Find dy/dx

Using the chain rule, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). Substituting the results from steps 1 and 2, we get: \(\frac{dy}{dx} = (-\sec u \tan u) \cdot (2x + 7)\).

Step 4: Substitute u Back Into the Expression

Substitute \(u = x^2 + 7x\) back into the expression derived in Step 3 so that the result is in terms of \(x\). We get \(\frac{dy}{dx} = -\sec(x^2 + 7x)\tan(x^2 + 7x)\cdot(2x + 7)\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. It's like peeling the layers of an onion, where each function is a layer inside another.
When you have two functions combined, such as \(y = f(u)\) and \(u = g(x)\), you want to find the derivative \(\frac{dy}{dx}\). The chain rule states that you multiply the derivative of the outer function by the derivative of the inner function.
In formula terms, it looks like this:

• First, find \(\frac{dy}{du}\) by differentiating the outer function \(f(u)\) with respect to its variable \(u\).
• Next, find \(\frac{du}{dx}\) by differentiating the inner function \(g(x)\) with respect to \(x\).
• Finally, multiply these derivatives together: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

This rule is extremely useful when dealing with nested functions, as it simplifies the process of differentiation and ensures you don't miss any steps.

Trigonometric Functions

Trigonometric functions like \(\sin\), \(\cos\), and \(\sec\) are crucial in calculus, especially when dealing with periodic or oscillating phenomena. These functions have special derivatives that you need to remember.

• The derivative of \(\sin u\) is \(\cos u\).
• The derivative of \(\cos u\) is \(-\sin u\).
• The derivative of \(\sec u\) is \(\sec u \tan u\).

These derivatives are helpful when you need to find the rate of change of these functions concerning another variable. In our example, we started with \(y = -\sec u\). To differentiate it with respect to \(u\), we used the known derivative of \(\sec u\), recalling it became \(\sec u \tan u\). Be aware and mindful of how the negative sign interacts, as it applies to the entire derivative in our problem.

Power Rule

The power rule is another vital tool in differentiation, especially when dealing with polynomials. It states that if \(u = x^n\), then the derivative \(\frac{du}{dx} = nx^{n-1}\). This rule is what you use when you see any power of \(x\) in a function.
For example, if you have \(u = x^2 + 7x\), you apply the power rule to each term separately:

• For the term \(x^2\), the derivative is \(2x\).
• For the term \(7x\), the derivative is \(7\).

Therefore, the derivative of \(u\) with respect to \(x\) becomes \(\frac{du}{dx} = 2x + 7\).
Understanding and applying the power rule helps simplify the differentiation of polynomial functions, making it quicker and easier to proceed with solving calculus problems. It's essential to break down each term, apply the differentiation separately, and then combine them for a complete derivative result.