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Chapter 1: Problem 14

Now find the derivative of each of the following functions. \(f(x)=x^{5} 5^{x}\)

Short Answer

Expert verified
The derivative of the function \(f(x)=x^{5} 5^{x}\) is \(f'(x) = 5x^{4} \cdot 5^{x} + x^{5} \cdot 5^{x} \ln(5)\)

Step by step solution

01

Identify the two functions in the product

Let's call the two functions in the product as \(u(x)\) and \(v(x)\) for ease of notation. Therefore, \(u(x) = x^{5}\) and \(v(x) = 5^{x}\).
02

Use the product rule to start the derivatization

According to the product rule, the derivative of \(f(x)\) is \(f'(x) = u'(x)v(x) + u(x)v'(x)\). The next step is to find \(u'(x)\) and \(v'(x)\)
03

Find the derivative of \(u(x)\)

The derivative of \(u(x) = x^{5}\) is straightforward, using the power rule for derivatives, \(u'(x) = 5x^{4}\).
04

Find the derivative of \(v(x)\)

The derivative of \(v(x) = 5^{x}\) is a bit trickier. The chain rule is required in order to find it. The chain rule states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function, which in this case yields \(v'(x) = 5^{x} \ln(5)\).
05

Compute the derivative \(f'(x)\)

Now that we have both \(u'(x)\) and \(v'(x)\), we can plug these into our formula from step 2. Therefore, \(f'(x) = u'(x)v(x) + u(x)v'(x) = 5x^{4} \cdot 5^{x} + x^{5} \cdot 5^{x} \ln(5)\).

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Key Concepts

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

Product Rule
In calculus, when you have a function that is the product of two or more simpler functions, you often need to find the derivative. This is where the **product rule** comes in handy. It helps us differentiate functions like \(f(x) = x^5 \cdot 5^x\). In this case:
  • \(u(x) = x^5\) describes our first function.
  • \(v(x) = 5^x\) represents our second function.
The product rule states that the derivative of a product \(u(x)v(x)\) is:\[f'(x) = u'(x)v(x) + u(x)v'(x)\] This means we take the derivative of the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second function. It's a systematic way to break down complex derivatives.
Chain Rule
The **chain rule** is a critical tool when dealing with derivatives of composite functions. A composite function is essentially a function within another function, like peeling an onion. To find the derivative, the chain rule tells us to take the derivative of the outer function and multiply it by the derivative of the inner function.In the function \(v(x) = 5^x\), where we want to find the derivative \(v'(x)\), we consider:
  • The outer function is the exponential function \(5^x\).
  • The inner function is the exponent \(x\).
Applying the chain rule, the derivative is:\[v'(x) = 5^x \ln(5)\] The natural logarithm \(\ln(5)\) comes from differentiating the exponential part. The chain rule simplifies our work by allowing us to tackle the derivative in parts.
Power Rule
The **power rule** is one of the most straightforward rules in calculus for finding derivatives. It is particularly useful when dealing with functions of the form \(x^n\), where \(n\) is any real number. The rule states that the derivative of \(x^n\) is:\[\frac{d}{dx} (x^n) = nx^{n-1}\]When applied to \(u(x) = x^5\), we get:
  • \(u'(x) = 5x^{5-1} = 5x^4\)
The power rule shows that you just have to bring the exponent down in front as a multiplier and subtract one from the exponent. It's a simple, powerful tool that makes differentiating power functions quick and easy.

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Most popular questions from this chapter

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