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suppose F(x)=f(x)g(2x). if f(1)=3, f'(1)=1, g(2)=2, and g'(2)=5. find f'(1).
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\[F'(x)=f'(x)g(2x)+2g'(2x)f(x)\] plug in the appropriate numbers to get your answer
i assume the question meant find \[F'(1)\]
yes it it did... sorry
which rule did u use to set it up btw?
Product rule?
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yes i used the product rule, plus the chain rule
because the derivative if \[g(2x)\] is \[2\g'(2x)\]
\[2g'(2x)\] i meant
that is how i got \[F'(x)=f'(x)g(2x)+2g'(2x)f(x)\]
so \[F'(1)=f'(1)g(2)+2g'(2)f(1)\] then plug in the numbers for the arithmetic
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you got this?
im ok for now. thanks!
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