Question

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

Short answer

Question: Determine the derivative of the function g(x) = \(\dfrac{(x+1)e^x}{x-2}\). Answer: The derivative of g(x) is given by \(g'(x) = \dfrac{e^x(x-2) + x(x+1)e^x - 3(x+1)e^x}{(x-2)^2}\).

Step-by-step solution

Recognize Parts of the Function#g(x)

In g(x), the function can be identified as follows: Numerator function: \((x+1)e^x\) Denominator function: \(x-2\) #Step 2: Apply the product rule to product functions in the numerator#

Using the Product Rule

To find the derivative of the numerator function, we use the product rule. Let's denote: Function 1: \(f(x)=x+1\) Function 2: \(h(x)=e^x\) Now we find the derivatives of these functions: \(f'(x)=1\) \(h'(x)=e^x\) Using the product rule, the derivative of \((x+1)e^x\) is: \((x+1)e^x = f(x)h(x)\) Therefore, the derivative will be: \((x+1)e^x\text' = f'(x)h(x) + f(x)h'(x)\) Plug in the previously derived values: \((x+1)e^x\text' = (1)(e^x) + (x+1)(e^x)\) #Step 3: Apply the quotient rule to find the derivative#

Using the Quotient Rule

Now, we have the derivative of the numerator \((x+1)e^x\text'\) and can use the quotient rule to find the derivative of the entire function g(x): \(g'(x) = \dfrac{((x+1)e^x\text')(x-2) - ((x+1)e^x)((x-2)\text')}{(x-2)^2}\) Plugging in the derivatives from the previous steps, we get: \(g'(x) = \dfrac{((1)(e^x)+(x+1)(e^x))(x-2) - ((x+1)e^x)(1)}{(x-2)^2}\) Now, simplify: \(g'(x) = \dfrac{(e^x+(x+1)e^x)(x-2) - (x+1)e^x}{(x-2)^2}\) \(g'(x) = \dfrac{e^x(x-2) + (x+1)e^x(x-2) - (x+1)e^x}{(x-2)^2}\) \(g'(x) = \dfrac{e^x(x-2) + x(x+1)e^x - 2(x+1)e^x - (x+1)e^x}{(x-2)^2}\) \(g'(x) = \dfrac{e^x(x-2) + x(x+1)e^x - 3(x+1)e^x}{(x-2)^2}\) This is the derivative of the given function g(x).

Key concepts

Product Rule

The Product Rule is a vital tool for finding the derivative of a product of two functions. If you have two functions, say \( f(x) \) and \( h(x) \), the product rule states that the derivative of their product \( f(x)h(x) \) is given by:\[ (f(x)h(x))' = f'(x)h(x) + f(x)h'(x) \]Let's break that down:

• Step 1: Derive \( f(x) \) to get \( f'(x) \).
• Step 2: Derive \( h(x) \) to get \( h'(x) \).
• Step 3: Apply the rule: Multiply the derivative of the first function by the second function and add it to the first function times the derivative of the second function.

For our problem, the functions \( f(x) = x+1 \) and \( h(x) = e^x \) are used. Their derivatives are \( f'(x) = 1 \) and \( h'(x) = e^x \). Plugging into the formula:\[ (x+1)e^x \text' = (1)(e^x) + (x+1)(e^x) \] This shows how the product rule helps handle more complex differentiation.

Quotient Rule

When dealing with a function that is the division of two functions, the Quotient Rule comes into play. If \( g(x) = \frac{u(x)}{v(x)} \), the derivative \( g'(x) \) is calculated as:\[ \left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]It might look complex, but it's straightforward once you understand:

• Step 1: Derive the numerator \( u(x) \) to get \( u'(x) \).
• Step 2: Derive the denominator \( v(x) \) to get \( v'(x) \).
• Step 3: Substitute these into the formula.
• Step 4: Compute and simplify the result.

For our specific function, \( u(x) = (x+1)e^x \) and \( v(x) = x-2 \). Applying the Quotient Rule:\[ g'(x) = \frac{((1)(e^x)+(x+1)(e^x))(x-2) - ((x+1)e^x)(1)}{(x-2)^2} \]This rule helps us navigate the complexities of derivatives involving fractions.

Exponential Function

Exponential functions have a unique property that makes them easy to differentiate. They are of the form \( e^x \), where \( e \) is the base of natural logarithms. The derivative of an exponential function is simply:\[ \frac{d}{dx}(e^x) = e^x \]This self-replicating nature makes their integration into calculus smooth:

• Exponential functions grow rapidly, illustrating many real-world phenomena from population growth to radioactive decay.
• In our problem, the exponential function \( e^x \) appears in the product \( (x+1)e^x \).
• The derivative \( (x+1)e^x \text' = (1)(e^x) + (x+1)(e^x) \) demonstrates the crucial role of the exponential function's derivative.

Understanding exponential functions, with their straightforward derivatives, allows us to tackle increasingly complex calculus challenges with ease.