The question is If g(x) + xsin(g(x))=x^2, and g(0)=pi find the derivative of g(0). I have no idea where to start. Thanks for any help.
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Originally Posted by HighPoint The question is If g(x) + xsin(g(x))=x^2, and g(0)=pi find the derivative of g(0). I have no idea where to start. Thanks for any help. Rewrite the equation as $\displaystyle g(x)=x^2-x\sin(g(x))$ Take the derivative on your own. This is the answer you can check Spoiler: $\displaystyle g'(x)=2x-(x\cos[g(x)]g'(x)+\sin(g(x)))$ Now plug in x=0 and use the fact that g(0)=pi.
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