Question

find the slope of the curve at the given point.
$8y^{8}+5x^{7}=3y + 10x$ at $(1,1)$
the slope of the curve $8y^{8}+5x^{7}=3y + 10x$ at $(1,1)$ is
(type a simplified fraction.)

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Using implicit differentiation:
For \(8y^{8}\), by the chain - rule \(\frac{d}{dx}(8y^{8})=8\times8y^{7}\frac{dy}{dx}=64y^{7}\frac{dy}{dx}\)
For \(5x^{7}\), \(\frac{d}{dx}(5x^{7}) = 35x^{6}\)
For \(3y\), by the chain - rule \(\frac{d}{dx}(3y)=3\frac{dy}{dx}\)
For \(10x\), \(\frac{d}{dx}(10x)=10\)

The derivative of the equation \(8y^{8}+5x^{7}=3y + 10x\) is:
\(64y^{7}\frac{dy}{dx}+35x^{6}=3\frac{dy}{dx}+10\)

Step2: Solve for \(\frac{dy}{dx}\)

Rearrange the terms to isolate \(\frac{dy}{dx}\):
\(64y^{7}\frac{dy}{dx}-3\frac{dy}{dx}=10 - 35x^{6}\)
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(64y^{7}-3)=10 - 35x^{6}\)
So, \(\frac{dy}{dx}=\frac{10 - 35x^{6}}{64y^{7}-3}\)

Step3: Substitute \(x = 1\) and \(y = 1\)

When \(x = 1\) and \(y = 1\):
\(\frac{dy}{dx}=\frac{10-35\times1^{6}}{64\times1^{7}-3}=\frac{10 - 35}{64 - 3}=\frac{-25}{61}\)

Answer:

\(\frac{-25}{61}\)