Question

question let y be defined implicitly by the equation 4x^{3}-5y^{4}-5x - 7y = 1. use implicit differentiation to evaluate \\(\frac{dy}{dx}\\) at the point (-1,0). submit an exact answer. provide your answer below:

Explanation:

Step1: Differentiate each term

Differentiate $4x^{3}-5y^{4}-5x - 7y=1$ term - by - term with respect to $x$.
The derivative of $4x^{3}$ with respect to $x$ is $12x^{2}$ (using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$).
The derivative of $-5y^{4}$ with respect to $x$ is $-20y^{3}\frac{dy}{dx}$ (using the chain - rule $\frac{d}{dx}(f(y))=f^{\prime}(y)\frac{dy}{dx}$).
The derivative of $-5x$ with respect to $x$ is $-5$.
The derivative of $-7y$ with respect to $x$ is $-7\frac{dy}{dx}$.
The derivative of the constant $1$ with respect to $x$ is $0$. So we have:
$12x^{2}-20y^{3}\frac{dy}{dx}-5 - 7\frac{dy}{dx}=0$.

Step2: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side:
$-20y^{3}\frac{dy}{dx}-7\frac{dy}{dx}=5 - 12x^{2}$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(-20y^{3}-7)=5 - 12x^{2}$.
Then $\frac{dy}{dx}=\frac{5 - 12x^{2}}{-20y^{3}-7}$.

Step3: Evaluate at the point $(-1,0)$

Substitute $x=-1$ and $y = 0$ into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{x=-1,y = 0}=\frac{5-12(-1)^{2}}{-20(0)^{3}-7}=\frac{5 - 12}{-7}=\frac{-7}{-7}=1$.

Answer:

$1$