Question
use implicit differentiation to find an equation of the tangent line to the graph at the given point.$$x+y-1=\ln \left(x^{2}+y^{2}\right), \quad(1,0)$$
Step 1
The given equation is $x+y-1=\ln \left(x^{2}+y^{2}\right)$. Using the chain rule and the fact that the derivative of $\ln u$ is $\frac{1}{u} \frac{du}{dx}$, we get $$ 1+\frac{dy}{dx}=\frac{2x+2y\frac{dy}{dx}}{x^{2}+y^{2}}. $$ Show more…
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