Suppose $y=\ln \left(x^{2}+y^{2}\right)$. Find:
\[
y^{\prime}(1)=
\]

Answer

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Answer

$\boxed{y'(1) = 2}$

Steps

Step 1 ：$y = \ln(x^2 + y^2)$

Step 2 ：Differentiate both sides with respect to x: $\frac{1}{y'} = \frac{2x}{x^2 + y^2} + \frac{2yy'}{x^2 + y^2}$

Step 3 ：Rearrange the terms: $y'(1 - \frac{2y^2}{x^2 + y^2}) = \frac{2x}{x^2 + y^2}$

Step 4 ：Solve for y': $y' = \frac{2x}{x^2 + y^2 + 2y^2}$

Step 5 ：Substitute x = 1 and y = \ln(1^2 + \ln^2(1)) = 0: $y'(1) = \frac{2*1}{1^2 + 0 + 2*0^2} = 2$

Step 6 ：$\boxed{y'(1) = 2}$