Question

How can I find the derivative of the equation: m(t)=(1/2)e^-t(sint-cost)+(1/2). An explanation for each step would be super helpful!

Answer 1

w.r.t t right?

Answer 2

\[m(t)=\frac{1}{2}e^{-t}(\sin(t)-\cos(t))+\frac{1}{2}\]?

Answer 3

\[m'(t)=[\frac{1}{2}e^{-t}(\sin(t)-\cos(t))]' \\ m'(t)=\frac{1}{2}[e^{-t}(\sin(t)-\cos(t))]'\]
do you know the product rule?

Answer 4

Why did you put another t at the very end of the equation? And did you leave the \[\frac{ 1 }{ 2 }\] because the derivative is 0?

Answer 5

And yes I do know the product rule. Would you just use it between the \[\frac{ 1 }{ 2 }\] and the rest of the equation?

Answer 6

Did I write the expression down wrong?

Answer 7

\[2 m'(t)=[e^{-t}(\sin(t)-\cos(t))]' \text{ going \to rewrite with the constant multiple } \\ \text{ on the other side so \it looks less confusing } \\ \text{ now we are going \to apply product rule \to the \right hand side } \\ 2 m'(t)=[e^{-t}]' (\sin(t)-\cos(t))+(e^{-t}) [\sin(t)-\cos(t)]'\]

Answer 8

now you just need to be able to evaluate:
\[(e^{-t})' \text{ and } (\sin(t)-\cos(t))'\]

Answer 9

Ok so then it's\[2m'(t)=e ^{-t}(\sin(t)-\cos(t))+ e ^{-t}(\cos(t)+sint(t))\]?

Answer 10

last part looks good
but the first part needs one little change

Answer 11

\[\frac{d}{dx}e^u =e^u \cdot \frac{du}{dx} \\ \text{ we have } u=-x \\ \frac{du}{dx}=-1 \\ \text{ so } \frac{d}{dx}e^{-x}=e^{-x} \cdot (-1)=-e^{-x}\]

Answer 12

\[2 m'(t)=e^{-t}(\cos(t)-\sin(t))+e^{-t}(\cos(t)+\sin(t)) \\ m'(t)=\frac{1}{2}(e^{-t}(\cos(t)-\sin(t))+e^{-t}(\cos(t)+\sin(t)))\]
now you combine like terms inside that big ( )

Answer 13

But wouldn't what you said before mean that the first \[e ^{-t} was -e ^{-t}\]?

Answer 14

yep I distribute the -1 part though

Answer 15

Oh ok. Thanks so much!!!

Answer 16

it doesn't matter like I could have waited to distribute the -1 part of the -e^(-t) thing

Answer 17

like because I'm going to distribute the e^(-t) as well

Answer 18

because I see like terms in this

Answer 19

\[2 m'(t)=e^{-t}(\cos(t)-\sin(t))+e^{-t}(\cos(t)+\sin(t)) \\ 2m'(t)=e^{-t} \cos(t)-e^{-t}\sin(t)+e^{-t} \cos(t)+e^{-t} \sin(t)\]
do you see like terms ?

Answer 20

Yes the \[-e ^{-t} and +e ^{-t}\] cancel out. Then you can add the 2 cos(t) together so its \[2e ^{-t}\cos t\] and then the 2 cancels out when you get m' alone by dividing by 2

Answer 21

\[2 m'(t)=2 e^{-t}\cos(t)\]
so
\[m'(t)=e^{-t} \cos(t)\]

Answer 22

isn't it cute how all of that turned into that little bit?

Answer 23

Anyways do you have any problems with any step

Answer 24

I did use constant rule that is (c)'=0
I did use constant multiple rule that is (cf)'=c(f)'
And finally used product rule that is (uv)'=v' u+u' v

Answer 25

Haha I don't know about cute, but I definitely understand it all now. Thank you so much!

Answer 26

\[m(t)=\frac{1}{2}e^{-t} (\sin(t)-\cos(t))+\frac{1}{2} \\ m'(t)=(\frac{1}{2}e^{-t}(\sin(t)-\cos(t))+\frac{1}{2})' \\ \text{ by sum rule we have } \\ m'(t)=(\frac{1}{2}e^{-t}(\sin(t)-\cos(t))'+(\frac{1}{2})' \\ \text{ by constant rule we have } \\ m'(t)=(\frac{1}{2}e^{-t}(\sin(t)-\cos(t))'+0 \\ \text{ by constant multiple rule } \\ m'(t)=\frac{1}{2}(e^{-t}(\sin(t)-\cos(t))' \\ m'(t)=\frac{1}{2}(-e^{-t}(\sin(t)-\cos(t))+e^{-t}(\cos(t)+\sin(t)) \\ m'(t)=\frac{1}{2}(-e^{-t}\sin(t)+e^{-t}\cos(t)+e^{-t}\cos(t)+e^{-t}\sin(t)) \\ m'(t)=\frac{1}{2}(2 e^{-t} \cos(t)) \\ m'(t)=e^{-t} \cos(t) \]
Just thought I would put all of our work in one post.