anonymous
  • anonymous
d/dt ( 2^(t)2sin^(-1)(2t+1) derivatives.
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@tkhunny please help!
tkhunny
  • tkhunny
Looks like a good "paying attention" problem. What have you tried? This? \(\dfrac{d}{dt}2^{t}\cdot 2\cdot\sin^{-1}(2t+1)\) Personally, I would rewrite it just a little. This? \(\dfrac{d}{dt}2^{t+1}\cdot\sin^{-1}(2t+1)\)
anonymous
  • anonymous
\[d/dt (2^t \sin ^{-1}(2t+1)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
product rule and chain rule?
tkhunny
  • tkhunny
Okay, so no extra 2 and there is no need or temptation to rewrite it. Great. You have the idea. Let's see what you get. Hint: \(\dfrac{d}{dt}2^{t} = 2^{t}\cdot\log(2)\)
anonymous
  • anonymous
mmm i have no clue.
tkhunny
  • tkhunny
False. You had a clue a minute ago. Get it back.
anonymous
  • anonymous
ok,
anonymous
  • anonymous
chain rule for each numbers?
tkhunny
  • tkhunny
Hint: \(\dfrac{d}{dt}\sin^{-1}(t) = \dfrac{1}{\sqrt{1-t^{2}}}\) That's about all you need.
tkhunny
  • tkhunny
Stop asking questions and start writing the derivative.
tkhunny
  • tkhunny
\(\dfrac{d}{dt}f(t)\cdot g(t) = f(t)\cdot g'(t) + g(t)\cdot f'(t)\) Go!
anonymous
  • anonymous
\[d/dt (2t+1)= (2)\]
tkhunny
  • tkhunny
That's part of it. Keep going.
anonymous
  • anonymous
product rules? but it does not make sense \[2^t \sin^{-1} (2t+1) \] three parts. the product has two/ f(x)g(x)
anonymous
  • anonymous
@tkhunny
tkhunny
  • tkhunny
If there where three, you could just group them and make pairs. There are only two. \(2^{t}\) \(\sin^{-1}(2t+1) = asin(2t+1)\)
anonymous
  • anonymous
oh.. i see.
anonymous
  • anonymous
\[2^t*(asin(2t+1)'+(asin(2t+1)(2^t)'\] ?
tkhunny
  • tkhunny
There's that question again. Move forward. Do NOT stop to ask every time you flinch! Keep going!
anonymous
  • anonymous
I want to make sure that i am doing right. so i dont need to rise off all my effort and time writting wrong equation
tkhunny
  • tkhunny
Right. I understand your motivation. I am also telling you that it is holding you back. You have far too much fear. Learn the right things. Go the right direction. Go with confidence. What's next?
anonymous
  • anonymous
\[(2^t)(1/\sqrt(1-t)*dx (2t+1)+asin(2t+1)(2^tlong(2)\]
tkhunny
  • tkhunny
So Close!! \(\dfrac{d}{dt}asin(t) = \dfrac{1}{\sqrt{1-t^{2}}}\) \(\dfrac{d}{dt}asin(2t+1) = \dfrac{1}{\sqrt{1-(2t+1)^{2}}}\cdot 2\)
tkhunny
  • tkhunny
See! That was very good. No need to fear. Just pay a little better attention.
anonymous
  • anonymous
\[2^t \frac{ 2 }{ \sqrt{1-(2t+1)}}+(\sin^{-1}(2t+1)(2^t* \log (2)) \]
tkhunny
  • tkhunny
Missed the "squared" in the square root. Otherwise, perfect.
anonymous
  • anonymous
sqrt(2t+1)?
tkhunny
  • tkhunny
Already showed you that. Pay better attention. \(\dfrac{2}{\sqrt{1 - (2t+1)^{2}}}\)
anonymous
  • anonymous
ok, thank you
tkhunny
  • tkhunny
Now, seriously. This is where I look you in the eye and ask you if you are in the right class, if you are truly dedicated to this course, if you think you can gain the confidence to move forward - and various other questions about your mental health. Work hard. You'll get it!
anonymous
  • anonymous
Thank you very much tkhunny. i will work hard and thank you for your advice.

Looking for something else?

Not the answer you are looking for? Search for more explanations.