2.2.211 AP Calculus Exam practice problem

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SUMMARY

The discussion centers on the calculus problem where \( y = x \sin x \) and the derivative \( \frac{dy}{dx} \) needs to be determined. The correct answer is option (B) \( \sin x + x \cos x \). Participants debated the notation, specifically whether to use \( y' \) alongside \( \frac{dy}{dx} \). Dan confirmed that if \( \frac{dy}{dx} \) is provided in the problem statement, it is best to maintain that notation for consistency.

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karush
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211(DOY)
If If $y=x \sin x,$ then $\dfrac{dy}{dx}=$
(A) $\sin x + \cos x$
(B) $\sin x + x \cos x$
(C) $\sin x + \cos x$
(D) $x(\sin x + \cos x)$
(E) $x(\sin x - \cos x)$
Solution
note: $y'=\dfrac{dy}{dx}$
Apply the Product Rule $$(f\cdot g)'=f\:'\cdot g+f\cdot g'$$
then
$$y'=x'\sin x+(\sin x)' x$$
simplify
$$y'=\sin x+x\cos x \quad (B)$$

ok this is a relatively simple problem but was wondering if $y'$ should be used in combination with $\dfrac{dy}{dx}=$
also any other typos ,, suggestion,, etc
 
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karush said:
211(DOY)
If If $y=x \sin x,$ then $\dfrac{dy}{dx}=$
(A) $\sin x + \cos x$
(B) $\sin x + x \cos x$
(C) $\sin x + \cos x$
(D) $x(\sin x + \cos x)$
(E) $x(\sin x - \cos x)$
Solution
note: $y'=\dfrac{dy}{dx}$
Apply the Product Rule $$(f\cdot g)'=f\:'\cdot g+f\cdot g'$$
then
$$y'=x'\sin x+(\sin x)' x$$
simplify
$$y'=\sin x+x\cos x \quad (B)$$

ok this is a relatively simple problem but was wondering if $y'$ should be used in combination with $\dfrac{dy}{dx}=$
also any other typos ,, suggestion,, etc
It looks good. If you are given [math]\dfrac{dy}{dx}[/math] in the problem statement then I would stick with that notation.

-Dan
 

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