Suppose $y = \ln (x^{2} + y^{2})$ . Find:
$y^{'} (1) =$

Answer

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Answer

$y^{'} (1) = 2$

Steps

Step 1 ： $y = \ln (x^{2} + y^{2})$

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

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

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

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 ： $y^{'} (1) = 2$