Question: Use implicit differentiation to find the slope of the tangent line to the curve
$y x - 9 y = x^6 - 7$
at the point $(1, \frac{6}{53})$.
slope =
Use implicit differentiation to determine $\frac{dy}{dx}$ given the equation $x^5 \cdot y^2 + \cos(x) = e^y$.
$\frac{dy}{dx} =$
1. **Problem 1:** Find the slope of the tangent line to the curve $y x - 9 y = x^6 - 7$ at the point $(1, \frac{6}{53})$ using implicit differentiation.
2. Differentiate both sides with respect to $x$. Remember $y$ is a function of $x$, so use the product rule and chain rule:
$$\frac{d}{dx}(y x) - \frac{d}{dx}(9 y) = \frac{d}{dx}(x^6 - 7)$$
3. Apply the product rule to $y x$:
$$\frac{d}{dx}(y x) = y \cdot \frac{d}{dx}(x) + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx}$$
4. Differentiate $9 y$:
$$\frac{d}{dx}(9 y) = 9 \frac{dy}{dx}$$
5. Differentiate the right side:
$$\frac{d}{dx}(x^6 - 7) = 6 x^5$$
6. Substitute these into the equation:
$$y + x \frac{dy}{dx} - 9 \frac{dy}{dx} = 6 x^5$$
7. Group terms with $\frac{dy}{dx}$:
$$x \frac{dy}{dx} - 9 \frac{dy}{dx} = 6 x^5 - y$$
8. Factor out $\frac{dy}{dx}$:
$$\frac{dy}{dx} (x - 9) = 6 x^5 - y$$
9. Solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{6 x^5 - y}{x - 9}$$
10. Evaluate at the point $(1, \frac{6}{53})$:
$$\frac{dy}{dx} = \frac{6 (1)^5 - \frac{6}{53}}{1 - 9} = \frac{6 - \frac{6}{53}}{-8} = \frac{\frac{318}{53} - \frac{6}{53}}{-8} = \frac{\frac{312}{53}}{-8} = \frac{312}{53} \cdot \frac{1}{-8} = -\frac{312}{424} = -\frac{39}{53}$$
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11. **Problem 2:** Use implicit differentiation to find $\frac{dy}{dx}$ given:
$$x^5 y^2 + \cos(x) = e^y$$
12. Differentiate both sides with respect to $x$:
$$\frac{d}{dx}(x^5 y^2) + \frac{d}{dx}(\cos(x)) = \frac{d}{dx}(e^y)$$
13. Use product rule on $x^5 y^2$:
$$\frac{d}{dx}(x^5 y^2) = 5 x^4 y^2 + x^5 \cdot 2 y \frac{dy}{dx} = 5 x^4 y^2 + 2 x^5 y \frac{dy}{dx}$$
14. Differentiate $\cos(x)$:
$$\frac{d}{dx}(\cos(x)) = -\sin(x)$$
15. Differentiate $e^y$ using chain rule:
$$\frac{d}{dx}(e^y) = e^y \frac{dy}{dx}$$
16. Substitute all derivatives:
$$5 x^4 y^2 + 2 x^5 y \frac{dy}{dx} - \sin(x) = e^y \frac{dy}{dx}$$
17. Group terms with $\frac{dy}{dx}$:
$$2 x^5 y \frac{dy}{dx} - e^y \frac{dy}{dx} = \sin(x) - 5 x^4 y^2$$
18. Factor out $\frac{dy}{dx}$:
$$\frac{dy}{dx} (2 x^5 y - e^y) = \sin(x) - 5 x^4 y^2$$
19. Solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{\sin(x) - 5 x^4 y^2}{2 x^5 y - e^y}$$