for the curve defined by the equation below, find $\frac{dy}{dx}$. $3y^{9}+2x^{6}y - x^{7}=1$ $\frac{dy}{dx}=$

Explanation:

Step1: Differentiate both sides

Differentiate $3y^{9}+2x^{6}y - x^{7}=1$ with respect to $x$ using product - rule and chain - rule. $\frac{d}{dx}(3y^{9})+\frac{d}{dx}(2x^{6}y)-\frac{d}{dx}(x^{7})=\frac{d}{dx}(1)$

Step2: Apply chain - rule to $3y^{9}$

By the chain - rule, $\frac{d}{dx}(3y^{9}) = 3\times9y^{8}\frac{dy}{dx}=27y^{8}\frac{dy}{dx}$.

Step3: Apply product - rule to $2x^{6}y$

The product - rule states that $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = 2x^{6}$ and $v = y$. So $\frac{d}{dx}(2x^{6}y)=2x^{6}\frac{dy}{dx}+12x^{5}y$.

Step4: Differentiate $x^{7}$ and $1$

$\frac{d}{dx}(x^{7}) = 7x^{6}$ and $\frac{d}{dx}(1)=0$.

Step5: Combine terms

$27y^{8}\frac{dy}{dx}+2x^{6}\frac{dy}{dx}+12x^{5}y - 7x^{6}=0$.

Step6: Isolate $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(27y^{8}+2x^{6})=7x^{6}-12x^{5}y$. Then $\frac{dy}{dx}=\frac{7x^{6}-12x^{5}y}{27y^{8}+2x^{6}}$.

Answer:

$\frac{7x^{6}-12x^{5}y}{27y^{8}+2x^{6}}$