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Use Version I of the Chain Rule to calculate \(\frac{d y}{d x}\). $$y=(3 x+7)^{10}$$

Short Answer

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Answer: \(\frac{d y}{d x} = 30 (3x+7)^9\)

Step by step solution

01

Identify the inner and outer functions

In the given function \(y=(3x+7)^{10}\), the inner function is \(u=3x+7\) and the outer function is \(v=u^{10}\).
02

Differentiate the inner function with respect to x

We need to find the derivative of the inner function (u) with respect to x: \(\frac{d u}{d x}\). Differentiating \(u=3x+7\), we get: $$\frac{d u}{d x} = 3$$
03

Differentiate the outer function with respect to the inner function

Now, we need to find the derivative of the outer function (v). We will differentiate \(v=u^{10}\) with respect to u: \(\frac{d v}{d u}\). Differentiating \(v=u^{10}\), we get: $$\frac{d v}{d u} = 10 u^9$$
04

Apply the Chain Rule

According to Version I of the Chain Rule, the derivative of the composite function \(y\) with respect to x is: $$\frac{d y}{d x} = \frac{d v}{d u} \cdot \frac{d u}{d x}$$
05

Substitute the derivatives

Now, substitute the derivatives we found in Steps 2 and 3 into the expression: $$\frac{d y}{d x} = (10 u^9) \cdot (3)$$
06

Replace u with its expression in terms of x

Since \(u=3x+7\), replace u in the derivative: $$\frac{d y}{d x} = 30 (3x+7)^9$$ So, the derivative of the function \(y=(3x+7)^{10}\) with respect to x using Version I of the Chain Rule is \(\frac{d y}{d x} = 30 (3x+7)^9\).

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Key Concepts

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

Composite Function Differentiation
When it comes to understanding calculus, one critical component is the concept of composite function differentiation. A composite function is created when one function is nested inside another; this occurs when the output of one function becomes the input of another. In simple terms, if you have a function z that takes in a value y, and y itself is a function that takes in a value x, then the function that directly relates z to x is a composite function.

To differentiate these functions, you cannot simply apply the basic rules of differentiation to the entire expression. Instead, you must consider the derivatives of the inner and outer functions separately in a process known as the Chain Rule. This rule is so important because it allows us to tackle complex derivatives of functions that are dependent on other functions. By breaking down these dependencies into their parts, we are using the concept of a composite function to understand the bigger picture.

In the provided exercise, the function \(y=(3x+7)^{10}\) clearly demonstrates a composite relationship, where the outer function \(y=u^{10}\) and the inner function \(u=3x+7\). This nesting of functions is precisely what the Chain Rule is designed to handle.
Finding Derivatives
To find the derivative of a function means to determine the rate at which its value changes with respect to changes in its input variable. The derivative of a function at a point is the slope of the tangent line to the function's graph at that point. While this concept may sound complex, it's akin to understanding the instantaneous speed of a car at any given moment.

Let's break down the process of finding derivatives into more palpable steps. First, you identify the type of function you're dealing with—whether it's a basic polynomial, trigonometric function, exponential, etc. Each type has its specific set of rules for differentiation. In our exercise, the function is a polynomial, and polynomials are generally straightforward to differentiate.

The exercise outlines these steps methodically, starting with the differentiation of the inner function \(u=3x+7\) resulting in a derivative of 3. The next step involves differentiating the outer function \(u^{10}\), which yields \(10u^9\). It's through these individual differentiations and the subsequent application of the Chain Rule that we can understand the derivative of the composite function as a whole.
Applying the Chain Rule
Applying the Chain Rule effectively is crucial for correctly finding the derivatives of composite functions. As its name suggests, this rule forms a 'chain' of derivatives when applied, linking the derivatives of the inner and outer functions. It's an invaluable tool in calculus and is integral to working through complex differentiation problems.

To use the Chain Rule, you take the derivative of the outer function, expressed as a function of the inner function, and then multiply it by the derivative of the inner function itself. Another way to think about this is to see the derivative as a rate of change. The Chain Rule tells us how to combine these rates of change when dealing with nested functions.

In the provided solution, after computing the derivatives of the inner and outer functions, we applied the Chain Rule: the derivative of the outer function (\(10u^9\)) is multiplied by the derivative of the inner function (3), giving us \(30(3x+7)^9\). This step is crucial because it connects all previous steps into the final derivative of y with respect to x. The elegance of the Chain Rule lies in its ability to simply complex differentiation into manageable parts that reflect the structure of the function we are examining.

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Most popular questions from this chapter

\(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that $$ \frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)} $$ b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0 ) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

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Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in \(\mathrm{km}\), the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}.\) a. Compute the pressure at the summit of Mt. Everest which has an elevation of roughly \(10 \mathrm{km}\). Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\). d. Does \(p^{\prime}(z)\) increase or decrease with \(z\) ? Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(\sin ^{-1} x\right)=\sqrt{1-x^{2}}$$

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