Question Statement

Given the function $y=(2\sqrt{x}+1)\left(x^3-6\right)$, find the slope of the tangent line at $x=0$.

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Background and Explanation

This problem requires the product rule for differentiation to find the derivative of a product of two functions. You will also need the power rule to handle the terms with fractional exponents (roots) and polynomial terms.

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Solution

We begin with the function: $y=(2\sqrt{x}+1)\left(x^3-6\right)$

To find the slope of the tangent at $x=0$, we must first compute $\frac{dy}{dx}$ and then evaluate it at $x=0$.

Finding the Derivative

Let $u=2\sqrt{x}+1=2x^{1/2}+1$ and $v=x^3-6$.

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

First, compute the individual derivatives:

• $\frac{du}{dx}=\frac{d}{dx}(2x^{1/2}+1)=2\cdot \frac{1}{2}{x}^{-1/2}+0=x^{-1/2}=\frac{1}{\sqrt{x}}$
• $\frac{dv}{dx}=\frac{d}{dx}(x^3-6)=3x^2$

Now apply the product rule:

$\begin{array}{rl}\frac{dy}{dx}& =\left(x^3-6\right)\frac{d}{dx}(2\sqrt{x}+1)+(2\sqrt{x}+1)\frac{d}{dx}\left(x^3-6\right)\\ & =\left(x^3-6\right)\cdot \left(2\cdot \frac{1}{2}(x{)}^{\frac{1}{2}-1}+0\right)+(2\sqrt{x}+1)\left(3x^2-0\right)\\ & =\left(x^3-6\right)\left(x^{-1/2}\right)+(2\sqrt{x}+1)3x^2\\ & =\left(x^3-6\right)\cdot \frac{1}{x^{1/2}}+(2\sqrt{x}+1)3x^2\end{array}$

Simplifying the Expression

Combine the terms over a common denominator of $x^{1/2}$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{\left(x^3-6\right)+x^{1/2}(2\sqrt{x}+1)3x^2}{x^{1/2}}\\ & =\frac{\left(x^3-6\right)+\left(2x+x^{1/2}\right)3x^2}{x^{1/2}}\text{[since }2\sqrt{x}\cdot x^{1/2}=2x\text{]}\\ & =\frac{x^3-6+6x^3+3x^{5/2}}{x^{1/2}}\\ \frac{dy}{dx}& =\frac{7x^3+3x^{5/2}-6}{x^{1/2}}\end{array}$

Evaluating at $x=0$

Substitute $x=0$ into the derivative:

$\frac{dy}{dx}=\frac{7(0{)}^3+3(0{)}^{5/2}-6}{(0{)}^{1/2}}=\frac{0+0-6}{0}=\infty$

Since the denominator is zero while the numerator is non-zero ($-6$):

• Slope of tangent $=\infty$ (undefined)
• This indicates the tangent line is vertical at $x=0$

Therefore, the slope does not exist in the finite sense.

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[f(x)\cdot g(x)]=f(x)\cdot g^{\prime }(x)+g(x)\cdot f^{\prime }(x)$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Derivative of a constant: $\frac{d}{dx}(c)=0$
• Exponent conversion: $\sqrt{x}=x^{1/2}$

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Summary of Steps

1. Identify the two functions: $u=2\sqrt{x}+1$ and $v=x^3-6$
2. Differentiate each function using the power rule: $\frac{du}{dx}=\frac{1}{\sqrt{x}}$ and $\frac{dv}{dx}=3x^2$
3. Apply the product rule: $\frac{dy}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}$
4. Simplify by finding a common denominator of $x^{1/2}$ and combining like terms
5. Substitute $x=0$ into the final expression $\frac{7x^3+3x^{5/2}-6}{x^{1/2}}$
6. Interpret the undefined result (division by zero) as an infinite slope, indicating a vertical tangent line