Question

a. use implicit differentiation to find $\frac{dy}{dx}$. b. find the slope of the curve at the given point. $cos y = 2x^{3}-2; (1,\frac{pi}{2})$ a. $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $\cos y$ with respect to $x$ using chain - rule and $2x^{3}-2$ with respect to $x$.
$-\sin y\frac{dy}{dx}=6x^{2}$

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

Isolate $\frac{dy}{dx}$ by dividing both sides by $-\sin y$.
$\frac{dy}{dx}=-\frac{6x^{2}}{\sin y}$

Step3: Find the slope at the given point

Substitute $x = 1$ and $y=\frac{\pi}{2}$ into $\frac{dy}{dx}$.
$\sin(\frac{\pi}{2}) = 1$, so $\frac{dy}{dx}\big|_{x = 1,y=\frac{\pi}{2}}=-\frac{6\times1^{2}}{1}=-6$

Answer:

a. $\frac{dy}{dx}=-\frac{6x^{2}}{\sin y}$
b. $-6$