f(x)=2xand g(x)=2^x
Find the composite function of fg(x)
fg(x)
=f(g(x))
=f(2^x)
=2(2^x)
I don’t understand how this in turn equals to 2^(x+1)
[Moderator's note: Moved from a technical forum and thus no template.]
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206 (day of year number)
If $f(x)=\sin{(\ln{(2x)})}$, then $f'(x)=$
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(B) $\dfrac{\cos{(\ln{(2x)}}}{x}$
(C) $\dfrac{\cos{(\ln{(2x)}}}{2x}$
(D) $\cos{\left(\dfrac{1}{2x}\right)}$
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and
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Proof Required.
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Many Thanks
John.
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[/B]
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The questions is asking me to find \frac{f}{g} basically , the question is asking me to find the answer , even though i know it, i can't get my head around it.
the composite function is
f(x)=x^2+1
g(x)=1/x
we need to find foG (f of g) [composite functions].
Homework Statement
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