Jdare123
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I'm having trouble answering this question. Im not sure how to answer it and the steps to answer it. So far this is my attempt but I'm not sure.

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mqb2766
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(Original post by Jdare123)
I'm having trouble answering this question. Im not sure how to answer it and the steps to answer it. So far this is my attempt but I'm not sure.

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How do you get the x(h-1)/h by removing g()?
Can you spot what type of functions you're dealing with?
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Jdare123
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(Original post by mqb2766)
How do you get the x(h-1)/h by removing g()?
Can you spot what type of functions you're dealing with?
Im not sure. Would i need to use the product rule on g(x)g(h) or some other process to answer that part?
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mqb2766
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(Original post by Jdare123)
Im not sure. Would i need to use the chain rule on g(x)g(h) or some other process to answer that part?
You can factor g(x) out, then reason (use the given information) about the remaining expression which only deprnds on h (and g(h)).
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Jdare123
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(Original post by Jdare123)
Im not sure. Would i need to use the product rule on g(x)g(h) or some other process to answer that part?
is some thing like this correct? Name:  95483238_248723226539636_5637209155128262656_n.jpg
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mqb2766
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(Original post by Jdare123)
is some thing like this correct? Name:  95483238_248723226539636_5637209155128262656_n.jpg
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Dont think so.
Youve got
(g(x)g(h) - g(x))/h
Factorise the numeratlor. One term depends on x, the other on h.

Did you think about what the function might be? Just think about its basic property: addition becomes multiplication.
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RDKGames
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(Original post by Jdare123)
is some thing like this correct?
Nope.

\displaystyle \begin{aligned} g'(x) & = \displaystyle \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h} \\[0.3cm] &= \lim_{h \to 0} \dfrac{g(x)g(h) - g(x)}{h} \\[0.3cm] &= g(x)\lim_{h \to 0}\dfrac{g(h) - 1}{h} \\[0.3cm] &= g(x) \lim_{h \to 0} \dfrac{g(0+h) - g(0)}{h}\end{aligned}

Finish it off... That final limit represents something.. what is it?
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