Question

Derivatives Find and simplify the derivative of the following functions. $$h(x)=\frac{x e^{x}}{x+1}$$

Short answer

Question: Determine and simplify the derivative of the function $$h(x)=\frac{x e^{x}}{x+1}$$. Answer: The derivative of the given function is $$h'(x) = \frac{e^x(2x+1)+x^2e^{x}}{(x+1)^2}$$.

Step-by-step solution

Writing the Quotient Rule formula

To differentiate $$h(x)=\frac{x e^{x}}{x+1}$$, we'll use the quotient rule, which states that if we have a function $$f(x)=\frac{g(x)}{h(x)}$$, then its derivative is $$f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}$$ For our function, $$g(x)=x e^{x}$$ and $$h(x)=x+1$$.

Differentiating g(x) and h(x)

Before we can apply the quotient rule, we need to differentiate the functions $$g(x)$$ and $$h(x)$$ individually. For the function $$g(x)=xe^{x}$$, we need to use the product rule. If the function is in the form of $$u(x)v(x)$$, then the derivative is $$f'(x)=u'(x)v(x)+u(x)v'(x)$$. We have $$u(x)=x$$ and $$v(x)=e^{x}$$, and their derivatives are: $$u'(x)=1$$ $$v'(x)=e^{x}$$ Applying the product rule, we have: $$g'(x) = u'(x)v(x)+u(x)v'(x)=1(e^{x})+x(e^{x})=(1+x)e^{x}$$ Now, let's differentiate $$h(x)=x+1$$: $$h'(x) =1$$

Applying the Quotient Rule

Now we can apply the quotient rule using the expressions and their derivatives: $$h'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}$$

Plugging in the expressions

Now plug in the expressions we found in Steps 2 and 3: $$h'(x) = \frac{(1+x)e^{x}(x+1)-x e^{x}(1)}{(x+1)^2}$$

Simplifying the result

Now, we simplify the result: $$h'(x) = \frac{(e^{x}+xe^{x})(x+1)-x e^{x}}{(x+1)^2}$$ Multiply through: $$h'(x) = \frac{e^{x}x+e^{x}+xe^{x}x+xe^{x}-x e^{x}}{(x+1)^2}$$ Notice that we have two similar terms, specifically $$xe^{x}$$ from within the numerator. Combine the terms and simplify the result: $$h'(x) = \frac{e^x(2x+1)+xe^{x}x}{(x+1)^2}$$ Our final simplified derivative of the function is: $$h'(x) = \frac{e^x(2x+1)+x^2e^{x}}{(x+1)^2}$$

Key concepts

Quotient Rule

When you're trying to find the derivative of a function written as a fraction, like \( h(x) = \frac{x e^x}{x+1} \), you use the Quotient Rule. The Quotient Rule helps you handle the division of two functions efficiently.Here's how it works: If you have a function \( f(x) = \frac{g(x)}{h(x)} \), the derivative is given by:\[f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}\]The idea is to take the derivative of the top function \( g(x) \), multiply it by the bottom \( h(x) \), and then subtract the product of the original top \( g(x) \) with the derivative of the bottom \( h(x) \). Finally, divide the result by the square of the bottom function, \( (h(x))^2 \). This rule is great for tackling complicated fractions, ensuring you don't make mistakes in the process.

Product Rule

Within the process of using the Quotient Rule, you'll often encounter situations where a part of the function requires another specific rule: the Product Rule. This happens when a function you need to differentiate is a product of two other functions. For example, within our original function, the term \( g(x) = x e^x \) is a product.The Product Rule states that if you have two functions, \( u(x) \) and \( v(x) \), their derivative \( f'(x) \) is:\[f'(x) = u'(x)v(x) + u(x)v'(x)\]Here's how to apply it:

• Differentiate \( u(x) \) and call it \( u'(x) \).
• Keep \( v(x) \) as is and multiply \( u'(x)v(x) \).
• Keep \( u(x) \) the same and differentiate \( v(x) \) to get \( v'(x) \).
• Multiply \( u(x)v'(x) \).
• Add both parts together to finish the calculation.

For \( g(x) = x e^x \), let \( u(x) = x \) and \( v(x) = e^x \). We find that \( u'(x) = 1 \) and \( v'(x) = e^x \). Plugging these into the product rule, we get \( g'(x) = (1)(e^x) + (x)(e^x) = (1 + x)e^x \). Simple, yet powerful!

Simplification

Once the derivative is calculated with rules like the Quotient and Product Rules, your work is only halfway done. Simplification is the process of taking what looks like a complex expression and making it tidy and manageable.In our example, the derivative \( h'(x) = \frac{(1+x)e^{x}(x+1)-x e^{x}}{(x+1)^2} \) needed careful handling to simplify. The main goal is to combine like terms, factor where possible, and ensure the expression is straightforward.Here’s a glimpse into simplifying:

• Look for common factors. For instance, \( e^x \) is common and can be factored out.
• Combine similar terms, especially those that often appear multiple times, like \( x \cdot e^x \).
• Ensure the final expression doesn't have unnecessary complexities and is easy to interpret.

After simplification, the streamlined derivative is expressed as:\[h'(x) = \frac{e^x(2x+1) + x^2e^x}{(x+1)^2}\]This makes it clearer and easier to work with, especially if you're using the derivative for further calculations or analysis. Simplifying expressions is crucial to understanding and using derivatives effectively.