Question

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

Answer 1

you should put parenthesis around stuff because im not sure if this all exponents to exponents

Answer 2

\[\Large\rm h(t)=3e^{-6t} \cdot t^{7} \cdot (t+6)^3\]This?

Answer 3

You'll have to apply your product rule,
it will look kinda complicated since you have three terms though:\[\Large\rm (uvw)'=u'vw+uv'w+uvw'\]

Answer 4

Or you can use logarithmic differentiation if that approach I mentioned sounds too rough

Answer 5

\[h(t)=3e ^{-6t}*t ^{7}* (t+6)^{3}\] is the equation
What I thought:
\[-18te ^{-6t}*7t ^{6}*3(t+6)^{2}+t\]

Answer 6

ok Ill use the product rule, give me just a moment, Ill try to solve it.

Answer 7

Setup would look something like that I suppose :)
Differentiating the blue stuff.

Answer 8

ahh woops i made a boo boo ;c
there we go
\[\rm h'(t)=\color{royalblue}{(3e^{-6t})'} \cdot t^{7} \cdot (t+6)^3\quad+\quad 3e^{-6t} \cdot \color{royalblue}{(t^{7})'} \cdot (t+6)^3\]\[\rm \qquad+\quad 3e^{-6t} \cdot t^{7} \cdot \color{royalblue}{((t+6)^3)'}\]

Answer 9

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.

Answer 10

opps last one, (3(t+6)^2)+t

Answer 11

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?

Answer 12

Derivative of t+6 = 1+0

Answer 13

Yes ofc, silly misstake.their answer tho is:
\[-6te^{-6t}*(t+6)^{2}*(3t ^{2}+13t-21)\]
HOW?!

Answer 14

We have to do some factoring :)
Books tends to over simplify, sall good, we can get there.

Answer 15

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?