



 
Hi Emily. When you implicitly differentiate with respect to x, you get y' on the left side obviously. You need to use the chain rule to differentiate the right side. [ln(x)]' = 1/x, so [ln(x^{2} + y^{2}]] ' = [x^{2} + y^{2}]' / (x^{2} + y^{2}) and so on. Eventually you complete the chain rule and have an equation involving y', x and y. Simply substitute at this point: x = 1, y = 0. Solve for y'. Hope this helps,  


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