Question

find the slope of the tangent line of the function defined implicitly by 8xy + x^5 = 9 + ln(x^4y^4) at (x,y)=(1,1). slope = number (3 significant figures)

Explanation:

Step1: Differentiate both sides

Differentiate $8xy + x^{5}=9+\ln(x^{4}y^{4})$ with respect to $x$.
For the left - hand side:
Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ for $8xy$ where $u = 8x$ and $v = y$, we get $8y + 8x\frac{dy}{dx}$, and the derivative of $x^{5}$ is $5x^{4}$. So the derivative of the left - hand side is $8y + 8x\frac{dy}{dx}+5x^{4}$.
For the right - hand side:
The derivative of a constant $9$ is $0$. Using the properties of logarithms $\ln(ab)=\ln a+\ln b$, we have $\ln(x^{4}y^{4})=\ln x^{4}+\ln y^{4}=4\ln x + 4\ln y$.
The derivative of $4\ln x$ is $\frac{4}{x}$ and the derivative of $4\ln y$ is $\frac{4}{y}\frac{dy}{dx}$. So the derivative of the right - hand side is $\frac{4}{x}+\frac{4}{y}\frac{dy}{dx}$.
So we have the equation $8y + 8x\frac{dy}{dx}+5x^{4}=0+\frac{4}{x}+\frac{4}{y}\frac{dy}{dx}$.

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

Rearrange the terms to isolate $\frac{dy}{dx}$:
$8x\frac{dy}{dx}-\frac{4}{y}\frac{dy}{dx}=\frac{4}{x}-8y - 5x^{4}$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(8x-\frac{4}{y})=\frac{4}{x}-8y - 5x^{4}$.
Then $\frac{dy}{dx}=\frac{\frac{4}{x}-8y - 5x^{4}}{8x-\frac{4}{y}}$.

Step3: Substitute $x = 1$ and $y = 1$

Substitute $x = 1$ and $y = 1$ into the formula for $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{\frac{4}{1}-8\times1 - 5\times1^{4}}{8\times1-\frac{4}{1}}=\frac{4 - 8 - 5}{8 - 4}=\frac{-9}{4}=- 2.25$.

Answer:

$-2.25$