Question:

Let $f(x) = \frac{{3x^3 - 5x^2 + 2x}}{{x^2 - x + 3}}$. Use the product and quotient rules to find the derivative of $f(x)$. Show all your work clearly and simplify the results.

Answer:

To find the derivative of the function $f(x) = \frac{{3x^3 - 5x^2 + 2x}}{{x^2 - x + 3}}$, we will use the product and quotient rules. Let's compute it step by step:

Step 1: Find the derivative of the numerator and denominator separately. Using the product rule, we have:

$\frac{{d}}{{dx}}(3x^3 - 5x^2 + 2x) = (3)(3x^2) + (-2)(5x) + (2)(1)$

Simplifying, we get: $9x^2 - 10x + 2$

Similarly, finding the derivative of the denominator $(x^2 - x + 3)$, we get:

$\frac{{d}}{{dx}}(x^2 - x + 3) = (2x - 1)$

Step 2: Apply the quotient rule. The quotient rule states that if $f(x) = \frac{{g(x)}}{{h(x)}}$, then $f'(x) = \frac{{g'(x) \cdot h(x) - g(x) \cdot h'(x)}}{{(h(x))^2}}$. Applying this to our function $f(x)$, we have:

$f'(x) = \frac{{(9x^2 - 10x + 2)(x^2 - x + 3) - (3x^3 - 5x^2 + 2x)(2x - 1)}}{{(x^2 - x + 3)^2}}$

Now let's simplify this expression.

Expanding the numerator:

$(9x^2 - 10x + 2)(x^2 - x + 3) = 9x^4 - 18x^3 + 7x^2 - 28x + 6$

$(3x^3 - 5x^2 + 2x)(2x - 1) = 6x^4 - 11x^3 + 3x^2 - 2x$

Substituting these values back into the expression:

$f'(x) = \frac{{(9x^4 - 18x^3 + 7x^2 - 28x + 6) - (6x^4 - 11x^3 + 3x^2 - 2x)}}{{(x^2 - x + 3)^2}}$

Simplifying further:

$f'(x) = \frac{{9x^4 - 18x^3 + 7x^2 - 28x + 6 - 6x^4 + 11x^3 - 3x^2 + 2x}}{{(x^2 - x + 3)^2}}$

Combining like terms:

$f'(x) = \frac{{3x^4 - 7x^3 + 4x^2 - 26x + 6}}{{(x^2 - x + 3)^2}}$

Therefore, the derivative of $f(x)$ is $\frac{{3x^4 - 7x^3 + 4x^2 - 26x + 6}}{{(x^2 - x + 3)^2}}$.