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Mathematics 16 Online
geerky42 (geerky42):

\[y = \dfrac{10}{1+x^2}\] \[\text{Find }y'\]

hartnn (hartnn):

know chain rule ?

geerky42 (geerky42):

Yeah, but how can I apply it?

hartnn (hartnn):

the question is 10/(1+x^2) right ?

geerky42 (geerky42):

Wait, let f(x) = 10/x , g(x) = 1+x² so y' = f'(g(x))g'(x) Right?

geerky42 (geerky42):

Why are there so many viewers? lol.

hartnn (hartnn):

y= 10/(1+x^2) ? if yes, f(x) = 1/x g(x) = 1+x^2

hartnn (hartnn):

10 is constant, can be taken out...

geerky42 (geerky42):

Ok, thanks.

hartnn (hartnn):

welcome ^_^

OpenStudy (anonymous):

y=10/(x^2 + 1)

OpenStudy (anonymous):

y=10/(x^2 + 1)

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