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Calculus1 19 Online
OpenStudy (anonymous):

Find g'(x) for g(x)= e^(x)/(1+x^(2)) What rule do you use here?

OpenStudy (jhannybean):

\[g(x) = \frac{e^x}{1+x^2}\implies e^x(1+x^2)^{-1}\] Depends on how you want to solve it,eitehr product rule with what I transformed the function to, or quotient rule,

OpenStudy (perl):

you can use quotient rule

OpenStudy (anonymous):

ok so I got this for the product rule e^(x)*(1+x^(2))^(-1)+e^(x) and how do you find the derivative for (1+x^(2))^(-1)?

OpenStudy (jhannybean):

\[\frac{d}{dx} x^n = nx^{n-1}\]

OpenStudy (jhannybean):

\[f'(x) = f'g+g'f \implies e^x(1+x^2)^{-1} -1e^x(1+x^2)^{-2}(2x)\]

OpenStudy (jhannybean):

Write it so that you have all positive variables.\[f'(x)=\frac{e^x}{(1+x^2)^2}-\frac{2xe^x}{(1+x^2)^2}\]\[f'(x)=\frac{e^x(1-2x)}{(1+x^2)^2}\]

OpenStudy (anonymous):

ok I understand how you got that now. Thank you :)

OpenStudy (jhannybean):

No problemo :)

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