I have no idea how to start these problems. How does one get the original function given the derivative of the function? I would appreciate it if someone would guide me in doing problem 7 at least so I understand what I'm doing. Thank you!
3 Answers
$\begingroup$Hint. These problems seem (to me) to ask you to find antiderivatives after you have learned the rules for differentiation but before you have studied integration.
That suggests that you look at the product rule$$ (fg)'(x) = f'(x)g(x) + f(x)g'(x) $$and the chain rule$$ \frac{d}{dx}f(g(x)) = f'(g(x))g'(x) $$(and the quotient rule and the rule for sums) and try to match $f$ and $g$ to functions whose derivatives you know: powers and trigonometric functions.
At this point in your study there are no rules for doing these problems. Later on you will have more strategies.
$\endgroup$ $\begingroup$You should be able to recognize the form of each of these expressions as the result of some specific derivative rules.
For example, if I showed you the expression $$f'(x)g(x) + f(x)g'(x),$$ wouldn't you recognize it as the result of the "product rule" for finding $[f(x)g(x)]'$?
So now, you need to try to think of which of the "derivative rules" is illustrated by each of these, then try to notice which piece plays which role in the rule.
$\endgroup$ 3 $\begingroup$For #8, $f'(x)= \frac{(3x^2+ 5)sec(x)- (x^3+ 5x)sec(x)tan(x)}{sec^2(x)}$. The first thing I notice is that $3x^2+ 5$ is the derivative of $x^3+ 5x$ and that $sec(x)tan(x)$ is the derivative of $sec(x)$. In other words, with $p(x)= x^3+ 5x$ and $q(x)= sec(x)$, the numerator is p'(x)q(x)- p(x)q'(x). And then notice the denominator is $sec^2(x)= q^2$. AHA! Quotient rule! $f(x)= \frac{p(x)}{q(x)}$.
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