To find the equation of the tangent line to the graph of the function defined implicitly by $x + y - 1 = \ln(x^7 + y^7)$ at the point $(1, 0)$, we will use implicit differentiation.

Step 1: Differentiate both sides with respect to $x$

Given: $x + y - 1 = \ln(x^7 + y^7)$

Differentiate both sides with respect to $x$: $\frac{d}{dx}(x) + \frac{d}{dx}(y) - \frac{d}{dx}(1) = \frac{d}{dx}\left(\ln(x^7 + y^7)\right)$

This simplifies to: $1 + \frac{dy}{dx} = \frac{1}{x^7 + y^7} \cdot \left(7x^6 + 7y^6 \cdot \frac{dy}{dx}\right)$

Step 2: Simplify and Solve for $\frac{dy}{dx}$

Rewrite this as: $1 + \frac{dy}{dx} = \frac{7x^6 + 7y^6 \frac{dy}{dx}}{x^7 + y^7}$

Multiply both sides by $x^7 + y^7$ to eliminate the fraction: $(x^7 + y^7)(1 + \frac{dy}{dx}) = 7x^6 + 7y^6 \frac{dy}{dx}$

Expanding, we get: $x^7 + y^7 + \frac{dy}{dx}(x^7 + y^7) = 7x^6 + 7y^6 \frac{dy}{dx}$

Collect all terms involving $\frac{dy}{dx}$ on one side: $\frac{dy}{dx} (x^7 + y^7 - 7y^6) = 7x^6 - x^7 - y^7$

Thus, $\frac{dy}{dx} = \frac{7x^6 - x^7 - y^7}{x^7 + y^7 - 7y^6}$

Step 3: Evaluate $\frac{dy}{dx}$ at the Point $(1, 0)$

Substitute $x = 1$ and $y = 0$ into the derivative:

$\frac{dy}{dx} = \frac{7(1)^6 - (1)^7 - (0)^7}{(1)^7 + (0)^7 - 7(0)^6} = \frac{7 - 1 - 0}{1 + 0 - 0} = 6$

Therefore, the slope of the tangent line at $(1, 0)$ is $6$.

Step 4: Write the Equation of the Tangent Line

The tangent line at $(1, 0)$ with slope $6$ is: $y - 0 = 6(x - 1)$

Simplifying: $y = 6x - 6$

Final Answer

The equation of the tangent line to the graph at $(1, 0)$ is: $y = 6x - 6$

Would you like more details on this solution, or do you have questions?

Here are some related questions you might explore:

1. What is implicit differentiation and why is it useful in finding tangent lines?
2. How can you verify the correctness of the derivative found through implicit differentiation?
3. How does the tangent line relate to the concept of local linear approximation?
4. What happens to the tangent line equation if the function is modified slightly?
5. How does implicit differentiation apply in higher dimensions?

Tip: When using implicit differentiation, carefully collect terms involving $\frac{dy}{dx}$ on one side to isolate and solve for the derivative.