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mlddmlnog
Group Title
let f and g be differentiable functions such that
f(1)=2 f'(1)=3 f'(2)=4
g(1)=2 g'(1)=3 g'(2)=5
if h(x)=f(g(x)), then h'(1)=?
 one year ago
 one year ago
mlddmlnog Group Title
let f and g be differentiable functions such that f(1)=2 f'(1)=3 f'(2)=4 g(1)=2 g'(1)=3 g'(2)=5 if h(x)=f(g(x)), then h'(1)=?
 one year ago
 one year ago

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mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
the answer is 12.
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
So you have to kind of remember the DEFINITION of the chain rule for this one.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large h(x)=f(g(x))\]The chain rule will produce this.\[\large h'(x)=f'(g(x))\cdot g'(x)\]
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
oh.. wait. i'll try to do it myself now. thank you for giving me a start! :)
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
yay i got it :)
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
here i the next one that's kind of similar to this. but i don't get. If \[f(x)=x ^{3}+3x ^{2}+4x+5\] and g(x)=5, then g(f(x))=?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Hmmmmm....
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
The answer is 5.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
So here is how a composition of functions works. Everywhere you see an x, you replace it with f(x). Example:\[\large \color{orangered}{f(x)=2x}\]\[\large g(\color{cornflowerblue}{x})=\color{cornflowerblue}{x}+3\] \[\large g(\color{orangered}{f(x)})=\color{orangered}{f(x)}+3\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
In the problem we've been given, the function \(g(x)\) is CONSTANT. There are no x's! So when we plug f(x) into it, it should give us the same answer, because g(x) is always 5. Always constant. You could do the composition thing I explained earlier and it might make sense. If you try to plug in f(x) for any x's in g, you'll see that you have nowhere to actually plug it in.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Is your teacher any good? Because so far I'm really really disliking these problems. None of them are straight forward. It just feels like he gave you a list of puzzles to work on.
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
haha yes. she loves these types of questions. _____ all of her questions are like this. making our brains explode. it literally takes me like 8 hours to do hw. it's insane.
 one year ago
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