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Chapter 3: Problem 27

Find the derivative of the following functions. $$g(x)=\frac{e^{x}}{x^{2}-1}$$

Short Answer

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Question: Find the derivative of the function $$g(x)=\frac{e^x}{x^2-1}$$. Answer: The derivative of the function g(x) is $$g'(x) = \frac{e^x(x^2 - 2x - 1)}{(x^2 - 1)^2}$$.

Step by step solution

01

Calculate u'(x) and v'(x)

For our function, we need to find the derivatives of $$u(x) = e^x$$ and $$v(x) = x^2 - 1$$. These can be found as follows: $$u'(x) = \frac{d}{dx}(e^x) = e^x$$ Here, the derivative of an exponential function is just the function itself. $$v'(x) = \frac{d}{dx}(x^2 - 1) = 2x$$ Now that we have the derivatives of u(x) and v(x), we will apply the quotient rule to find the derivative of g(x).
02

Apply the quotient rule

Using the quotient rule, we can now find the derivative of the function g(x): $$ \begin{aligned} g'(x) &= \frac{d}{dx} \left(\frac{e^x}{x^2 - 1}\right) \\ &= \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \\ &= \frac{(e^x)(x^2 - 1) - (e^x)(2x)}{(x^2 - 1)^2} \end{aligned} $$
03

Simplify the expression

Now, we need to simplify the expression for g'(x): $$ \begin{aligned} g'(x) &= \frac{e^x(x^2 - 1) - e^x(2x)}{(x^2 - 1)^2} \\ &= \frac{e^x(x^2 - 1 - 2x)}{(x^2 - 1)^2} \end{aligned} $$ The final derivative of the function g(x) is: $$g'(x) = \frac{e^x(x^2 - 2x - 1)}{(x^2 - 1)^2}$$

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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 method used to find the derivative of a function that is the ratio of two differentiable functions. It is especially handy when you have a function as a fraction where both the numerator (top part) and the denominator (bottom part) are different functions. For a function of the form \( h(x) = \frac{f(x)}{g(x)} \), the derivative is given by:
  • \( h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \)
This specific rule requires you to:
  • Find the derivative of the numerator (\( f'(x) \))
  • Find the derivative of the denominator (\( g'(x) \))
  • Substitute into the formula \( \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \)
For the function \( g(x) = \frac{e^x}{x^2 - 1} \), we treated \( e^x \) as the numerator and \( x^2 - 1 \) as the denominator. We calculated their derivatives and substituted them into the quotient rule formula to find the derivative \( g'(x) \).
This method highlights the importance of calculating each part separately to avoid mistakes and ensure accuracy.
Exponential Function
Exponential functions are a type of function where the variable appears in the exponent. The most common base for exponential functions is the natural exponent \( e \), which is approximately equal to 2.718. The unique feature of exponential functions, particularly \( e^x \), is that their own derivative is the same as the original function. In other words, if \( u(x) = e^x \), then \( u'(x) = e^x \) as well.
This property makes them incredibly useful in calculus because it simplifies the differentiation process. There鈥檚 no need to manipulate the function further when differentiating; it remains unchanged. Therefore, when using exponential functions in derivative problems, especially in conjunction with other rules like the quotient rule,
the calculation becomes more straightforward as the exponential term does not change.
Simplifying Expressions
Simplifying expressions is a key step in mathematics, especially when working with derivatives. Once you calculate the derivative using rules like the quotient rule, you often end up with complex or lengthy expressions. Simplifying these expressions can help make sense of the result and may be necessary for further applications.When simplifying:
  • Combine like terms to reduce expressions.
  • Factor out common factors if possible to make the expression clearer.
  • Ensure the final expression is in the simplest form, so it can be easily interpreted and used.
With the function \( g'(x) \) resulting from our example, simplifying involved combining similar components of the expression \( e^x(x^2 - 1 - 2x) \) over \( (x^2 - 1)^2 \). Here, simplifying not only made the expression clearer but was essential to present the derivative in its neatest form.

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Most popular questions from this chapter

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