How do you find the derivative of (t-2)^3 (t-6) ?

Question

jhannybean · Answer

1 possible way is to expand $\ (t-2)^3$, multiply it into (t-6) and then take the derivative of the expansion.

anonymous · Answer

just expand out the whole thing to make it easier
x`(t)=4*t^3-30*t^2+72*t-54 
x`(t)=(2*(2*t-3))*(t-3)^2

jhannybean · Answer

$$\ (t-2)^3 = t^3-6t^2 +12t - 8 $$$$\ (t-2)^3(t-6) = t^4-12t^3+48t^2-80t+48$$

jhannybean · Answer

And then just take the derivative of this  $$\frac{d}{dt} (t^4-12t^3+48t^2-80t+48)$$

jhannybean · Answer

And you know how to take derivatives, right?

anonymous · Answer

yeah I'm just kind of new at this. I'm used to taking derivatives of let's say (t-2)^3 the extra (t-6) just throws me off. The chain rule is not needed right?

jhannybean · Answer

Nope, you can just take the derivative of the expansion as i showed above.

jhannybean · Answer

$$\ nt^{n-1}$$

unklerhaukus · Answer

$$y(t)=(t-2)^3 (t-6)$$
The product rule
$$y'(t)=\frac{\mathrm d}{\mathrm dx}\Big((t-2)^3\Big)(t-6)+(t-2)^3\frac{\mathrm d}{\mathrm dx}\Big(t-6\Big)$$

Now use the cain rule to find $\frac{\mathrm d}{\mathrm dx}\Big((t-2)^3\Big)$

unklerhaukus · Answer

chain*

jhannybean · Answer

Oh that's another method! Nice.

unklerhaukus · Answer

The expansion method should give an equivalent result,
but I think the product rule is a bit simpler for this question

jhannybean · Answer

Yeah, I agree.

anonymous · Answer

ok now I just have to stare at it to understand it. Thanks for your help :)

unklerhaukus · Answer

remember that the product rule is just $$y(x)=f(x)\cdot g(x)\
y'(x)=f'g+fg'$$
and the chain rule is$$y(x)=f(g(x))\y'(x)=f'(g(x))\cdot g'$$

mathstudent55 · Answer

Product rule & chain rule.

mathstudent55 · Answer

$y = uv$

$y' = uv' + vu'$

$y = (t - 2)^3(t - 6) $

$y' = (t - 2)^3 (t - 6)' + (t - 6)[(t - 2)^3]'$

$y' =(t - 2)^3 (1) + (t - 6)[3(t - 2)^2(1)]$

$y' = (t - 2)^3 + 3(t -6)(t - 2)^2$

$y' = (t - 2)^2 [(t - 2) + 3(t - 6)] $

$y'= (t - 2)^2[t - 2 + 3t - 18]$

$y' = (t - 2)^2 (4t -20)$

$y' = (t - 2)^2 (4)(t - 5)$

$y' = 4(t - 2)^2(t - 5) $

anonymous · Answer

ok so is the derivative of (t-2)^3 this --> 3(t-2)^2 (1)

mathstudent55 · Answer

Yes.

jhannybean · Answer

Yup

anonymous · Answer

ok I'm confused before I came here I asked this question on yahoo answers and someone said the derivative was 3(t-2)^2 (t-6)+ (t-2)^3 . Is it the same thing as the answer on here?

jhannybean · Answer

they half-used the product rule. It is incorrect.

anonymous · Answer

oh ok thank you so much for your help :D I finally understand it

mathstudent55 · Answer

@endoftime14
Notice that the answer you got from yahoo is the same as the 6th line of my answer above (with the order of the terms switched). The answer is correct, but I just simplified it more.