MHB 215 AP Calculus Exam Problem Power Rule

[karush]

If $y=(x^3-cos x)^5$, then $y'=$(A) $\quad 5(x^3-\cos x)^4$(B) $\quad 5(3x^2+\sin x)^4$(C) $\quad 5(3x^2+\sin x)^4$(D) $\quad 5(x^3+\sin x)^4(6x+\cos x)$(E) $\quad 5(x^3+\cos x)^4(3x+\sin x)$

Spoiler

The Power Rule

$\displaystyle{d\over dx}x^n = nx^{n-1}$

by observation we know the derivative of the inside is multiplied to the outside
thus this leaves option (E) the other steps probably are not necessary

ok I am sure this could be worded better. but I think many students take these tests and are not used multiple choice and just plunge into timely calculations when observation could be within a few seconds

 

[Greg]

Spoiler

$$y'=5(x^3-\cos x)^4(3x^2+\sin x)$$

 

[skeeter]

If $y=(x^3-cos x)^5$, then $y'=$(A) $\quad 5(x^3-\cos x)^4$(B) $\quad 5(3x^2+\sin x)^4$(C) $\quad 5(3x^2+\sin x)^4$(D) $\quad 5(x^3+\sin x)^4(6x+\cos x)$(E) $\quad 5(x^3+\cos x)^4(3x+\sin x)$

Spoiler

The Power Rule

$\displaystyle{d\over dx}x^n = nx^{n-1}$

by observation we know the derivative of the inside is multiplied to the outside
thus this leaves option (E) the other steps probably are not necessary

ok I am sure this could be worded better. but I think many students take these tests and are not used multiple choice and just plunge into timely calculations when observation could be within a few seconds

Choice (E) is closest to being correct, but as you typed it, has a sign error in the first variable factor and is missing a square in the second.
It's not just the power rule alone, it's also an application of the chain rule.

$\dfrac{d}{dx} u^n = nu^{n-1} \cdot \dfrac{du}{dx}$

$y = (x^3-\cos{x})^5 \implies y'=5(x^3-\cos{x})^4 (3x^2 + \sin{x})$

 

[karush]

If $y=(x^3-cos x)^5$, then $y'=$

(A) $\quad 5(x^3-\cos x)^4$

(B) $\quad 5(3x^2+\sin x)^4$

(C) $\quad 5(3x^2+\sin x)^4$

(D) $\quad 5(x^3+\sin x)^4(6x+\cos x)$

(E) $\quad 5(x^3-\cos x)^4(3x+\sin x)$

Spoiler

The ChainRule
$\displaystyle \dfrac{d}{dx}u^n = nu^{n-1} \cdot \dfrac{du}{dx}$
so
$y = (x^3-\cos{x})^5 \implies y'=5(x^3-\cos{x})^4 (3x^2 + \sin{x})\quad (E)$

hopefully

 

[skeeter]

If $y=(x^3-cos x)^5$, then $y'=$

(A) $\quad 5(x^3-\cos x)^4$

(B) $\quad 5(3x^2+\sin x)^4$

(C) $\quad 5(3x^2+\sin x)^4$

(D) $\quad 5(x^3+\sin x)^4(6x+\cos x)$

(E) $\quad 5(x^3-\cos x)^4(3x+\sin x)$

Spoiler

The ChainRule
$\displaystyle \dfrac{d}{dx}u^n = nu^{n-1} \cdot \dfrac{du}{dx}$
so
$y = (x^3-\cos{x})^5 \implies y'=5(x^3-\cos{x})^4 (3x^2 + \sin{x})\quad (E)$

hopefully

you forgot the square in the $\dfrac{du}{dx}$ factor ... again.

 

[karush]

where ?

 

[topsquark]

where ?

Look at answer key for E). It should be a [math]3x^2[/math], not 3x.

-Dan

 

[karush]

got it
thanks