Question Statement

Find $\frac{dy}{dx}$ given:

$y=\left(\frac{1}{x}+\frac{1}{x^2}\right)\left(3x^3+27\right)$

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Background and Explanation

This problem involves differentiating a product of two functions. You can approach this by either applying the product rule directly to the original expression, or by first simplifying the algebraic expression into a single rational function and then using the quotient rule (or power rule on individual terms).

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Solution

Method 1: Using the Product Rule

We identify the function as a product of $u=\left(\frac{1}{x}+\frac{1}{x^2}\right)$ and $v=\left(3x^3+27\right)$.

Applying the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$:

$\begin{array}{rl}\frac{dy}{dx}& =\left(3x^3+27\right)\frac{d}{dx}\left(\frac{1}{x}+\frac{1}{x^2}\right)+\left(\frac{1}{x}+\frac{1}{x^2}\right)\frac{d}{dx}\left(3x^3+27\right)\\ & =\left(3x^3+27\right)\left(\frac{d}{dx}\left(x^{-1}\right)+\frac{d}{dx}\left(x^{-2}\right)\right)+\left(\frac{1}{x}+\frac{1}{x^2}\right)\left(9x^2+0\right)\\ & =\left(3x^3+27\right)\left((-1)x^{-2}+\left(-2x^{-3}\right)\right)+\left(\frac{x+1}{x^2}\right)\left(9x^2\right)\\ & =\left(3x^3+27\right)\left(\frac{-1}{x^2}-\frac{2}{x^3}\right)+9x+9\\ & =\left(3x^3+27\right)\left(\frac{-x-2}{x^3}\right)+\frac{9x+9}{1}\\ & =\frac{-3x^4-6x^3-27x-54}{x^3}+\frac{9x+9}{1}\\ & =\frac{-3x^4-6x^3-27x-54+9x^4+9x^3}{x^3}\\ \frac{dy}{dx}& =\frac{6x^4+3x^3-27x-54}{x^3}\end{array}$

Method 2: Simplification First

First, combine the terms in the first factor and expand the product:

$\begin{array}{rl}y& =\left(\frac{1}{x}+\frac{1}{x^2}\right)\left(3x^3+27\right)\\ & =\left(\frac{x+1}{x^2}\right)\left(3x^3+27\right)\\ & =\frac{3x^4+27x+3x^3+27}{x^2}\\ & =\frac{3x^4+3x^3+27x+27}{x^2}\end{array}$

Now apply the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$ where $u=3x^4+3x^3+27x+27$ and $v=x^2$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{x^2\cdot \frac{d}{dx}\left(3x^4+3x^3+27x+27\right)-\left(3x^4+3x^3+27x+27\right)\frac{d}{dx}\left(x^2\right)}{(x^2{)}^2}\\ & =\frac{x^2\cdot \left(12x^3+9x^2+27+0\right)-\left(3x^4+3x^3+27x+27\right)(2x)}{x^4}\\ & =\frac{x\left[x\left(12x^3+9x^2+27\right)-\left(3x^4+3x^3+27x+27\right)2\right]}{x^4}\\ & =\frac{12x^4+9x^3+27x-6x^4-6x^3-54x-54}{x^3}\end{array}$

$\frac{dy}{dx}=\frac{6x^4+3x^3-27x-54}{x^3}$

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[u(x)v(x)]=u(x)\frac{dv}{dx}+v(x)\frac{du}{dx}$
• Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{v(x)\frac{du}{dx}-u(x)\frac{dv}{dx}}{[v(x){]}^2}$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$ (used for negative and fractional exponents)
• Algebraic simplification: Combining fractions $\frac{1}{x}+\frac{1}{x^2}=\frac{x+1}{x^2}$ before differentiating

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Summary of Steps

1. Identify the structure: Recognize $y$ as a product of two functions
2. Method 1 (Product Rule):
   • Differentiate each factor separately using the power rule
   • Apply the product rule formula: $u^{\prime }v+u{v}^{\prime }$
   • Find common denominators to combine terms into a single fraction
   • Simplify the numerator by combining like terms
3. Method 2 (Simplification):
   • Combine $\frac{1}{x}+\frac{1}{x^2}$ into $\frac{x+1}{x^2}$
   • Expand the numerator $(x+1)(3x^3+27)$
   • Apply the quotient rule to the resulting rational function
   • Factor and cancel common terms, then combine like terms
4. Final Result: Both methods yield $\frac{dy}{dx}=\frac{6x^4+3x^3-27x-54}{x^3}$