Problem Restated:

You need to find the slope of the tangent line to the curve $x^2 - 3xy + y^3 = 49$ at the point $(-2, 3)$. However, you are on Planet Z, where the power rule for differentiation is modified as:

$\frac{d}{dx} \, x^n = n^2 \, x^{n-1}.$

This twist will affect the derivative calculations involving terms like $x^2$, so let's carefully proceed step-by-step.

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1. Use Implicit Differentiation

The curve is:

$x^2 - 3xy + y^3 = 49.$

Since $y$ is a function of $x$ (i.e., $y = y(x)$), we need to differentiate both sides with respect to $x$, applying implicit differentiation.

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2. Differentiate Each Term

Using the Planet Z power rule, let’s differentiate term by term.

(a) Differentiate $x^2$

$\frac{d}{dx} \, (x^2) = 2^2 \, x^{2-1} = 4x.$

(b) Differentiate $-3xy$

We need to apply the product rule here:
$\frac{d}{dx} \, (-3xy) = -3 \left( \frac{d}{dx}(x) \cdot y + x \cdot \frac{dy}{dx} \right).$

• $\frac{d}{dx}(x) = 1$, so the first part becomes $-3 \cdot y = -3y$.
• The second part is $-3x \cdot \frac{dy}{dx} = -3x \cdot y'$, where $y' = \frac{dy}{dx}$.

Thus, the derivative of $-3xy$ is: $-3y - 3x \cdot y'.$

(c) Differentiate $y^3$

Using the chain rule and treating $y$ as a function of $x$: $\frac{d}{dx}(y^3) = 3^2 \, y^{3-1} \cdot y' = 9y^2 \cdot y'.$

(d) Differentiate the constant $49$

Since the derivative of a constant is $0$, we get: $\frac{d}{dx}(49) = 0.$

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3. Combine All Terms

Now, put everything together:

$4x - 3y - 3x \cdot y' + 9y^2 \cdot y' = 0.$

Rearrange terms so that all the $y'$ terms are on the right:

$-3x \cdot y' + 9y^2 \cdot y' = 3y - 4x.$

Factor out $y'$ on the left:

$y' \left( 9y^2 - 3x \right) = 3y - 4x.$

Solve for $y'$:

$y' = \frac{3y - 4x}{9y^2 - 3x}.$

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4. Evaluate the Derivative at the Point $(-2, 3)$

We need to find the slope of the tangent line at the point $(-2, 3)$. Substitute $x = -2$ and $y = 3$ into the derivative:

$y' = \frac{3(3) - 4(-2)}{9(3)^2 - 3(-2)}.$

Simplify the numerator:

$3(3) - 4(-2) = 9 + 8 = 17.$

Simplify the denominator:

$9(9) - 3(-2) = 81 + 6 = 87.$

Thus, the slope of the tangent line is:

$y' = \frac{17}{87}.$

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5. Final Answer

The slope of the tangent line to the curve at the point $(-2, 3)$ is:

$\boxed{\frac{17}{87}}.$

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Do You Have Any Questions?

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5 Follow-Up Questions:

1. What is implicit differentiation, and why is it used here?
2. How does the modified power rule change standard calculus operations?
3. Can you derive the product rule on Planet Z using the twisted power rule?
4. How would the slope change if the differentiation rules for $y$ were also modified?
5. What is the geometric significance of the slope of the tangent line?

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One Tip:

When using implicit differentiation, always remember to treat $y$ as a function of $x$, and apply the product or chain rule where necessary!