Chapter 3: Problem 48
Derivatives Find and simplify the derivative of the following functions. $$h(x)=\frac{x+1}{x^{2} e^{x}}$$
Short Answer
Expert verified
Answer: The simplified derivative of the function h(x) is $$h'(x)=\frac{-(x^{3} + 2x^{2} + 2x)}{x^{4} e^{x}}$$.
Step by step solution
01
Identify the rules to apply
In this exercise, we need to find the derivative of the function $$h(x)=\frac{x+1}{x^{2} e^{x}}$$. This function is a quotient of two functions, so we need to apply the quotient rule.
02
Apply the quotient rule and chain rule
The quotient rule states that if we have a function $$h(x)=\frac{f(x)}{g(x)}$$, then its derivative is $$h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$. In this case, our functions are:
$$f(x)=x+1$$
$$g(x)=x^{2} e^{x}$$
First, we find the derivatives of f(x) and g(x):
$$f'(x)=1$$
Using the product rule and chain rule for g(x):
$$g'(x) = (2x)e^{x} + x^{2}(e^{x}) = e^{x}(2x + x^{2})$$
Now we can apply the quotient rule to find the derivative of h(x):
$$h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}=\frac{1 \cdot x^{2} e^{x}-(x+1) e^{x} (2x + x^{2})}{(x^{2} e^{x})^{2}}$$
03
Simplify the expression
Now we simplify the expression for the derivative:
$$h'(x)=\frac{x^{2} e^{x}-2x^{2} e^{x} - x^{3} e^{x} - x e^{x}(2+x)}{x^{4} e^{2x}}$$
Combine like terms and factor out the common terms:
$$h'(x)=\frac{e^{x}(-x^{3} - 3x^{2} - 2x + x^{2})}{x^{4} e^{2x}}$$
Now factor out a -1 to simplify further:
$$h'(x)=\frac{-e^{x}(x^{3} + 3x^{2} + 2x - x^{2})}{x^{4} e^{2x}}$$
Combine like terms:
$$h'(x)=\frac{-e^{x}(x^{3} + 2x^{2} + 2x)}{x^{4} e^{2x}}$$
Now, divide both numerator and denominator by $$e^{x}$$ to simplify the expression:
$$h'(x)=\frac{-(x^{3} + 2x^{2} + 2x)}{x^{4} e^{x}}$$
So, the simplified derivative of the function h(x) is:
$$h'(x)=\frac{-(x^{3} + 2x^{2} + 2x)}{x^{4} e^{x}}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The Quotient Rule is a powerful tool in calculus used to find the derivative of a ratio of two functions. Consider a function \( h(x) = \frac{f(x)}{g(x)} \). The derivative of this function, according to the Quotient Rule, is given by:\[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]In simpler terms, the derivative of the quotient \( h(x) \) involves:
- First, finding the derivative of the numerator \( f'(x) \),
- Then, finding the derivative of the denominator \( g'(x) \),
- Subsequently, subtracting the products of these derivatives: \( f'(x)g(x) - f(x)g'(x) \),
- Finally, dividing the whole expression by the square of the denominator \( [g(x)]^2 \).
Product Rule
The Product Rule is employed when you need the derivative of a product of two functions. Suppose we have two functions \( u(x) \) and \( v(x) \) such that their product is \( y(x) = u(x)v(x) \). The derivative, using the Product Rule, is:\[ y'(x) = u'(x)v(x) + u(x)v'(x) \]Here's how this formula is structured:
- First, compute the derivative of the first function \( u'(x) \), then multiply it by the original form of the second function \( v(x) \).
- Second, compute the derivative of the second function \( v'(x) \), then multiply it by the original form of the first function \( u(x) \).
- Finally, add these results together.
Chain Rule
The Chain Rule is key when dealing with composite functions, where one function is nested inside another. If you have a function of the form \( F(x) = f(g(x)) \), then the derivative is:\[ F'(x) = f'(g(x)) \, g'(x) \]This means:
- First, find the derivative of the outer function \( f \) with respect to its argument \( g(x) \), \( f'(g(x)) \).
- Then, multiply this result by the derivative of the inner function \( g'(x) \).