Question

1. find an equation of the tangent line to the curve $ln(x + y)=8xy$ at the point $(1,0)$.

Explanation:

Step1: Differentiate both sides implicitly

Differentiate $\ln(x + y)$ with respect to $x$ using the chain - rule. The derivative of $\ln(u)$ is $\frac{u'}{u}$, where $u=x + y$, so $\frac{d}{dx}\ln(x + y)=\frac{1 + y'}{x + y}$. Differentiate $8xy$ with respect to $x$ using the product - rule $(uv)'=u'v+uv'$, where $u = 8x$ and $v = y$. So $\frac{d}{dx}(8xy)=8y+8xy'$. Then we have the equation $\frac{1 + y'}{x + y}=8y + 8xy'$.

Step2: Substitute the point $(1,0)$

Substitute $x = 1$ and $y = 0$ into the equation $\frac{1 + y'}{x + y}=8y + 8xy'$. We get $\frac{1 + y'}{1+0}=8\times0+8\times1\times y'$, which simplifies to $1 + y'=8y'$.

Step3: Solve for $y'$

Subtract $y'$ from both sides of the equation $1 + y'=8y'$: $1=8y'-y'$, so $7y' = 1$ and $y'=\frac{1}{7}$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,0)$ and $m = y'=\frac{1}{7}$. Substituting these values, we get $y-0=\frac{1}{7}(x - 1)$.

Answer:

$y=\frac{1}{7}(x - 1)$ or $y=\frac{1}{7}x-\frac{1}{7}$