anonymous
  • anonymous
derrivate h(t)=3e^-6t*t^7*(t+6)^3 I thought it would be -18te^-6t*7t^6*3(t+6)^2+t PLEASE HELP
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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sweetburger
  • sweetburger
you should put parenthesis around stuff because im not sure if this all exponents to exponents
zepdrix
  • zepdrix
\[\Large\rm h(t)=3e^{-6t} \cdot t^{7} \cdot (t+6)^3\]This?
zepdrix
  • zepdrix
You'll have to apply your product rule, it will look kinda complicated since you have three terms though:\[\Large\rm (uvw)'=u'vw+uv'w+uvw'\]

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zepdrix
  • zepdrix
Or you can use logarithmic differentiation if that approach I mentioned sounds too rough
anonymous
  • anonymous
\[h(t)=3e ^{-6t}*t ^{7}* (t+6)^{3}\] is the equation What I thought: \[-18te ^{-6t}*7t ^{6}*3(t+6)^{2}+t\]
anonymous
  • anonymous
ok Ill use the product rule, give me just a moment, Ill try to solve it.
zepdrix
  • zepdrix
Setup would look something like that I suppose :) Differentiating the blue stuff.
zepdrix
  • zepdrix
ahh woops i made a boo boo ;c there we go \[\rm h'(t)=\color{royalblue}{(3e^{-6t})'} \cdot t^{7} \cdot (t+6)^3\quad+\quad 3e^{-6t} \cdot \color{royalblue}{(t^{7})'} \cdot (t+6)^3\]\[\rm \qquad+\quad 3e^{-6t} \cdot t^{7} \cdot \color{royalblue}{((t+6)^3)'}\]
anonymous
  • anonymous
yup I did exactly that can I'll just write the blue on every hope u understand. The first blue would be -18e^-6t, second: 7t^6. last: 3(t+6)+t. I just putthem together with the black, but that answer doesnt match the one in my book, why they gotta always mess things up....I follow their rules and stuff they never follow their own.
anonymous
  • anonymous
opps last one, (3(t+6)^2)+t
zepdrix
  • zepdrix
woops, last blue should give us this:\[\Large\rm \color{royalblue}{\frac{d}{dx}(t+6)^3}=3(t+6)^2\color{royalblue}{(t+6)'}=3(t+6)(1)\]ya?
zepdrix
  • zepdrix
Derivative of t+6 = 1+0
anonymous
  • anonymous
Yes ofc, silly misstake.their answer tho is: \[-6te^{-6t}*(t+6)^{2}*(3t ^{2}+13t-21)\] HOW?!
zepdrix
  • zepdrix
We have to do some factoring :) Books tends to over simplify, sall good, we can get there.
zepdrix
  • zepdrix
So what does everything have? They each have the exponential, ya? How many t's do they all have? Well they all have at least 6 t's multiplying. t^6 And they all have a factor of 3, right? :d See where we're going with this?
zepdrix
  • zepdrix
|dw:1434882566785:dw|
zepdrix
  • zepdrix
We want to pull out the greatest common factor that each term shares.
zepdrix
  • zepdrix
|dw:1434882683506:dw|So we pull that out.