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Chapter 3: Problem 33

Use Version 2 of the Chain Rule to calculate the derivatives of the following functions. $$y=(\sec x+\tan x)^{5}$$

Short Answer

Expert verified
Question: Compute the derivative of the function $$y=(\sec x+\tan x)^{5}$$ using the Chain Rule. Answer: The derivative of the given function is $$\frac{dy}{dx} = 5(\sec x+\tan x)^4(\sec x\tan x +\sec^{2} x)$$.

Step by step solution

01

Identify the outer and inner functions

We have the composite function $$y=(\sec x+\tan x)^{5}$$. Let's define the outer function as $$v(u)=u^5$$ and inner function as $$u(x)=\sec x+\tan x$$. We need to find $$\frac{dy}{dx}$$.
02

Calculate the derivative of the outer function regarding u

Now, we find the derivative of the outer function: $$\frac{dv}{du}$$. Using the power rule, we get: $$\frac{dv}{du}=5u^4$$.
03

Calculate the derivative of the inner function regarding x

Next, we find the derivative of the inner function: $$\frac{du}{dx}$$. Using the sum rule and recalling the derivatives for secant and tangent functions, we obtain: $$\frac{du}{dx}=\frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) = \sec x\tan x +\sec^{2} x$$.
04

Apply the Chain Rule to calculate the derivative of y concerning x

Finally, we will apply the Chain Rule, which states: $$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$$. We already have the values for $$\frac{dv}{du}$$ and $$\frac{du}{dx}$$ from Steps 2 and 3. Now, we substitute those values and the inner function $$u(x)$$ into the equation: $$\frac{dy}{dx} = \frac{dv}{du}\cdot\frac{du}{dx}= 5u^4\cdot (\sec x\tan x +\sec^{2} x)$$. Now we substitute $$u(x)=\sec x+\tan x$$ back into the equation to get the derivative of the original function y concerning x: $$\frac{dy}{dx} = 5(\sec x+\tan x)^4\cdot (\sec x\tan x +\sec^{2} x)$$. Hence, the derivative of the given function is: $$\frac{dy}{dx} = 5(\sec x+\tan x)^4(\sec x\tan x +\sec^{2} x)$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of Composite Functions
Understanding the derivative of composite functions is essential when dealing with more complex equations in calculus. Imagine putting on a pair of socks and then slipping into your shoes; the sock is the inner function, and the shoe is the outer function. In mathematical terms, if you have two functions, say f(g(x)), where f is the outer function and g is the inner function, the derivative is not simply the product of each function's derivative. Rather, it's like peeling the layers: start with the outer shell and work your way in.

To systematically approach this, first, take the derivative of the outer function as though the inner function were a simple variable. Then, multiply it by the derivative of the inner function. This process is known as the Chain Rule. Using this rule allows you to tackle functions within functions effectively and is particularly handy for our example, where y is a composite of a trigonometric and power function.
Power Rule
The Power Rule is a quick shortcut in differentiation. It's basically stating that to find the derivative of a variable raised to a power, you bring down the exponent as a multiplier and reduce the original exponent by one. In mathematical terms, if you have a function such as f(x) = x^n, the derivative f'(x) will be nx^(n-1).

Why does it matter? It simplifies the process of finding derivatives for monomials and comes in handy when dealing with polynomials too. In our exercise, we used the power rule for the outer function v(u) = u^5, which gave us a derivative of 5u^4. Always remember to acknowledge that x is being raised to a power in a function, a sign that the power rule is likely your friend.
Sum Rule
Moving on to the Sum Rule, which plays a role in our approach to tackling the derivative of the composite function. The sum rule in differentiation is the idea that the derivative of a sum of functions is the sum of the derivatives. This is a direct and straightforward rule, which states (f + g)' = f' + g'. This simplifies the entire process when you're dealing with functions that are being added together.

In practical use, when you encounter a function like u(x) = sec x + tan x, you would differentiate each function independently and then add the results. As we did in the exercise, we found the derivatives for secant and tangent functions separately and then applied the sum rule.
Trigonometric Function Derivatives
Finally, it's time to talk about the derivatives of trigonometric functions, which can be quite peculiar compared to algebraic functions. Each trigonometric function has its own derivative formula that you'll need to memorize for efficient calculus work. For instance, the derivative of sin(x) is cos(x), and that of cos(x) is -sin(x), but what about sec(x) and tan(x)?

It gets interesting: the derivative of sec(x) is sec(x)tan(x), while the derivative of tan(x) is sec^2(x). These trigonometric derivatives come into play when we step into the realm of the sum rule and the chain rule, just as they did in our exercise where we combined the power rule, the sum rule, and the knowledge of trigonometric derivatives to find the solution.

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Most popular questions from this chapter

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One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general, $$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\)

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