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Mathematics 18 Online
OpenStudy (anonymous):

suppose F(x)=f(x)g(2x). if f(1)=3, f'(1)=1, g(2)=2, and g'(2)=5. find f'(1).

OpenStudy (anonymous):

\[F'(x)=f'(x)g(2x)+2g'(2x)f(x)\] plug in the appropriate numbers to get your answer

OpenStudy (anonymous):

i assume the question meant find \[F'(1)\]

OpenStudy (anonymous):

yes it it did... sorry

OpenStudy (anonymous):

which rule did u use to set it up btw?

OpenStudy (anonymous):

Product rule?

OpenStudy (anonymous):

yes i used the product rule, plus the chain rule

OpenStudy (anonymous):

because the derivative if \[g(2x)\] is \[2\g'(2x)\]

OpenStudy (anonymous):

\[2g'(2x)\] i meant

OpenStudy (anonymous):

that is how i got \[F'(x)=f'(x)g(2x)+2g'(2x)f(x)\]

OpenStudy (anonymous):

so \[F'(1)=f'(1)g(2)+2g'(2)f(1)\] then plug in the numbers for the arithmetic

OpenStudy (anonymous):

you got this?

OpenStudy (anonymous):

im ok for now. thanks!

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