Question

practice exercises 13–26. implicit differentiation carry out the following steps. a. use implicit differentiation to find $\frac{dy}{dx}$. b. find the slope of the curve at the given point. 13. $x^{4}+y^{4}=2;(1, - 1)$ 14. $x = e^{y};(2,ln 2)$ 15. $y^{2}=4x;(1,2)$ 16. $y^{2}+3x = 8;(1,sqrt{5})$ 17. $sin y = 5x^{4}-5;(1,pi)$ 18. $sqrt{x}-2sqrt{y}=0;(4,1)$ 19. $cos y = x;(0,\frac{pi}{2})$ 20. $\tan xy = x + y;(0,0)$

Explanation:

Step1: Differentiate both sides of the equation $x^{4}+y^{4}=2$ with respect to $x$

Using the power - rule and chain - rule, we have:
$$\frac{d}{dx}(x^{4})+\frac{d}{dx}(y^{4})=\frac{d}{dx}(2)$$
$$4x^{3}+4y^{3}\frac{dy}{dx}=0$$

Step2: Solve for $\frac{dy}{dx}$

Subtract $4x^{3}$ from both sides:
$$4y^{3}\frac{dy}{dx}=-4x^{3}$$
Then divide both sides by $4y^{3}$:
$$\frac{dy}{dx}=-\frac{x^{3}}{y^{3}}$$

Step3: Find the slope at the point $(1, - 1)$

Substitute $x = 1$ and $y=-1$ into $\frac{dy}{dx}$:
$$\frac{dy}{dx}\big|_{(1,-1)}=-\frac{1^{3}}{(-1)^{3}} = 1$$

Answer:

a. $\frac{dy}{dx}=-\frac{x^{3}}{y^{3}}$
b. The slope of the curve at the point $(1,-1)$ is $1$.