You can ask questions about the Review for Quiz 2 by responding to this post. You can also answer said questions as well!
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Can we go over in class question 1.c me and a group of classmates cannot seem to figure it out? google is not helping :(.
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also, will there be no compound interest on the test? there is no questions about it on the review.
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You mean the derivative of
f(x) = \sin(x) - \cos(x) + \tan(x)?
The one thing that you don’t know just yet is the derivative of the tangent function; we’ll be talking about that in class on Wednesday.
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That is correct! Perhaps, it will show up on the next exam but I felt that we had plenty on the quiz already.
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How would you go after solving problem number four? That’s the question that asks us to prove that
\frac{d}{dx} (2f(x) - g(x)) = 2f'(x) - g'(x).
I’m having some trouble understanding what is expected for the quiz surrounding this question in particular.
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First off, I’m sorry I missed this question - I don’t know how that happened! At any rate, the question is still certainly still relevant for the next exam so … here’s my response:
The first critical issue is writing down the difference quotient for the function. Recall that, given any function (say F(x)), the difference quotient for F(x) is
\frac{F(x+h)-F(x)}{h}.
By definition, the limit as h\to0 of this thing is exactly F'(x). That is,
\lim_{h\to0} \frac{F(x+h)-F(x)}{h} = F'(x).
Now, in this problem, your function is 2f(x)-g(x); so we’ve got to write down the difference quotient for that. That works out to be
\frac{(2f(x+h)-g(x+h)) - (2f(x)-g(x))}{h}.
Now, a little algebra on that and we see that it’s the same as
2\frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}.
In that final expression, we see the difference quotient for f(x) and the difference quotient for g(x); those converge to f'(x) and g'(x) respectively.
Putting that altogether in one answer we get:
\begin{aligned}
\frac{d}{dx} (2f(x) - g(x)) &= \lim_{h\to0} \frac{F(x+h)-F(x)}{h} \\
&= \lim_{h\to0} \left(2\frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}\right) \\
&= 2f'(x) - g'(x).
\end{aligned}
To be clear, that string of equalities in those last three lines should be the full answer.
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