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1-44. Find the derivative of each function. $$ f(x)=x^{3} e^{x} $$

Short Answer

Expert verified
The derivative is \( e^x(x^3 + 3x^2) \).

Step by step solution

01

Identify the Structure

The function given is a product of two functions: \( x^3 \) and \( e^x \). This means we'll need to use the product rule to find the derivative.
02

Recall the Product Rule Formula

The product rule states that \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \). Here, \( u = x^3 \) and \( v = e^x \).
03

Differentiate the First Function

Find the derivative of \( u = x^3 \). The derivative \( u' = 3x^2 \).
04

Differentiate the Second Function

Find the derivative of \( v = e^x \). The derivative \( v' = e^x \), since the derivative of \( e^x \) with respect to \( x \) is \( e^x \).
05

Apply the Product Rule

Substitute \( u \, (x^3) \), \( u' \, (3x^2) \), \( v \, (e^x) \), and \( v' \, (e^x) \) into the product rule: \( (uv)' = u'v + uv' = 3x^2 e^x + x^3 e^x \).
06

Simplify the Expression

Factor out the common term \( e^x \) in the expression: \( = e^x(3x^2 + x^3) \), which simplifies the expression for the derivative.

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Key Concepts

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

Product Rule
The product rule is an essential tool in calculus used for differentiating functions that are multiplied together. Imagine you have two functions, say \( u(x) \) and \( v(x) \), that form a product \( uv \). To find the derivative of this product, we use the product rule. The formula is given by:
  • \((uv)' = u'v + uv'\)
This means you take the derivative of the first function \( u \), multiply it by the second function \( v \), and then add the product of the first function \( u \) with the derivative of the second function \( v \).

Using this rule ensures that you carefully handle how each function changes, considering both simultaneously. It particularly comes into play when dealing with more complex functions where simple rules cannot be directly applied. The product rule is possible due to the linearity and limit properties of derivatives.
Function Differentiation
Function differentiation refers to the process of finding the derivative of a function. A derivative provides us with the rate at which a function is changing at any given point. In simple terms, it tells us how steeply a function climbs or falls as it moves along its curve.

There are basic rules for differentiation, with each rule applied depending on the type of function you are dealing with. For example:
  • The power rule is used for polynomial functions, which states that \( \frac{d}{dx}x^n = nx^{n-1} \).
  • For trigonometric functions, periodic derivatives are used, such as \( \sin(x) \) and \( \cos(x) \).
In problems involving compounded functions like products or quotients, specialized rules like the product rule and the chain rule are applied. Understanding these rules helps analyze and simplify the differentiation process. Differentiation is an intrinsic part of calculus that enables solving complex real-world problems involving change.
Exponential Function Derivative
The exponential function \( e^x \) is a unique and vital function in mathematics due to its amazing property: its own derivative is itself! The function \( e^x \) represents continuous growth, such as population growth or capital investment with constant compounding. When you differentiate \( e^x \), you simply get \( e^x \) again.
  • \( \frac{d}{dx}e^x = e^x \)
This characteristic simplifies many calculations in calculus because it eliminates the need to alter the structure of the function when deriving.

Exponential functions frequently appear in physics, finance, and statistics, making their differentiation key to understanding growth patterns in these areas. For more complex algebraic operations involving the exponential function, rules such as the product and chain rules are employed to handle the derivation efficiently. Whether simple or compounded within other functions, the exponential function's consistent nature aids significantly in differentiation tasks.

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Most popular questions from this chapter

BIOMEDICAL: Diabetes The percentage of the U.S. population with diabetes is $$ D(x)=0.96+0.011 e^{x}+\ln x $$ where \(x\) is the number of decades since \(1950 .\) Find \(D^{\prime}\) (5) and interpret your answer.

97-98. ATHLETICS: World's Record 100-Meter Run In 1987 Carl Lewis set a new world's record of 9.93 seconds for the 100 -meter run. The distance that he ran in the first \(x\) seconds was $$ 11.274\left[x-1.06\left(1-e^{-x / 1.06}\right)\right] \text { meters } $$ for \(0 \leq x \leq 9.93 .\) Enter this function as \(y_{1},\) and define \(y_{2}\) as its derivative (using NDERIV), so that \(y_{2}\) gives his velocity after \(x\) seconds. Graph them on the window [0,9.93] by [0,100] Trace along the velocity curve to verify that Lewis's maximum speed was about 11.27 meters per second. Find how quickly he reached a speed of 10 meters per second, which is \(95 \%\) of his maximum speed.

PERSONAL FINANCE: Compound Interest A sum of \(\$ 1000\) at \(5 \%\) interest compounded continuously will grow to \(V(t)=1000 e^{0.05 t}\) dollars in \(t\) years. Find the rate of growth after: a. 0 years (the time of the original deposit). b. 10 years.

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\(55-58\). For each function: a. Find \(f^{\prime}(x)\) b. Evaluate the given expression and approximate it to three decimal places. $$ f(x)=\ln \left(e^{x}-1\right), \text { find and approximate } f^{\prime}(3) . $$

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