Question Statement

Differentiate the following function with respect to $x$:

$y=\frac{x-2}{x^4+x+1}$

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Background and Explanation

This problem involves differentiating a rational function (a quotient of two polynomials). When differentiating a function of the form $y=\frac{u}{v}$, we must use the quotient rule, which requires calculating the derivatives of both the numerator and denominator separately before combining them according to a specific formula.

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Solution

We begin by identifying the components of our function. Let $u=x-2$ (the numerator) and $v=x^4+x+1$ (the denominator).

First, calculate the derivatives of $u$ and $v$ with respect to $x$:

• $\frac{d}{dx}(x-2)=1$
• $\frac{d}{dx}(x^4+x+1)=4x^3+1$

Now apply the quotient rule formula: $\frac{dy}{dx}=\frac{v\cdot \frac{du}{dx}-u\cdot \frac{dv}{dx}}{v^2}$

Substituting our expressions:

$\frac{dy}{dx}=\frac{\left(x^4+x+1\right)\frac{d}{dx}(x-2)-(x-2)\frac{d}{dx}\left(x^4+x+1\right)}{{\left(x^4+x+1\right)}^2}$

Evaluate the derivatives in the numerator:

$=\frac{\left(x^4+x+1\right)(1-0)-(x-2)\left(4x^3+1+0\right)}{{\left(x^4+x+1\right)}^2}$

Simplify the terms:

$=\frac{x^4+x+1-(x-2)\left(4x^3+1\right)}{{\left(x^4+x+1\right)}^2}$

Expand the product $(x-2)(4x^3+1)=4x^4+x-8x^3-2$ in the numerator:

$=\frac{x^4+x+1-\left(4x^4+x-8x^3-2\right)}{{\left(x^4+x+1\right)}^2}$

Distribute the negative sign through the parentheses:

$=\frac{x^4+x+1-4x^4-x+8x^3+2}{{\left(x^4+x+1\right)}^2}$

Combine like terms in the numerator ($x^4-4x^4=-3x^4$, $x-x=0$, and $1+2=3$):

$=\frac{-3x^4+8x^3+3}{{\left(x^4+x+1\right)}^2}$

Therefore, the derivative is:

$\frac{dy}{dx}=\frac{-3x^4+8x^3+3}{(x^4+x+1{)}^2}$

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Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Sum/Difference Rule: $\frac{d}{dx}[f(x)\pm g(x)]=f^{\prime }(x)\pm g^{\prime }(x)$
• Constant Rule: $\frac{d}{dx}(c)=0$

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Summary of Steps

1. Identify components: Set $u=x-2$ and $v=x^4+x+1$
2. Differentiate separately: Find $\frac{du}{dx}=1$ and $\frac{dv}{dx}=4x^3+1$
3. Apply quotient rule: Substitute into $\frac{v(1)-u(4x^3+1)}{v^2}$
4. Expand: Multiply out $(x-2)(4x^3+1)=4x^4+x-8x^3-2$
5. Distribute negative: Change signs of all terms in the expanded product
6. Combine like terms: Group $x^4$ terms, $x$ terms, and constants to simplify the numerator to $-3x^4+8x^3+3$
7. Final form: Write the simplified derivative $\frac{-3x^4+8x^3+3}{(x^4+x+1{)}^2}$