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d/dt ( 2^(t)2sin^(-1)(2t+1) derivatives.

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@tkhunny please help!
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? \(\dfrac{d}{dt}2^{t+1}\cdot\sin^{-1}(2t+1)\)
\[d/dt (2^t \sin ^{-1}(2t+1)\]

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Other answers:

product rule and chain rule?
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: \(\dfrac{d}{dt}2^{t} = 2^{t}\cdot\log(2)\)
mmm i have no clue.
False. You had a clue a minute ago. Get it back.
chain rule for each numbers?
Hint: \(\dfrac{d}{dt}\sin^{-1}(t) = \dfrac{1}{\sqrt{1-t^{2}}}\) That's about all you need.
Stop asking questions and start writing the derivative.
\(\dfrac{d}{dt}f(t)\cdot g(t) = f(t)\cdot g'(t) + g(t)\cdot f'(t)\) Go!
\[d/dt (2t+1)= (2)\]
That's part of it. Keep going.
product rules? but it does not make sense \[2^t \sin^{-1} (2t+1) \] three parts. the product has two/ f(x)g(x)
If there where three, you could just group them and make pairs. There are only two. \(2^{t}\) \(\sin^{-1}(2t+1) = asin(2t+1)\)
oh.. i see.
\[2^t*(asin(2t+1)'+(asin(2t+1)(2^t)'\] ?
There's that question again. Move forward. Do NOT stop to ask every time you flinch! Keep going!
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
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?
\[(2^t)(1/\sqrt(1-t)*dx (2t+1)+asin(2t+1)(2^tlong(2)\]
So Close!! \(\dfrac{d}{dt}asin(t) = \dfrac{1}{\sqrt{1-t^{2}}}\) \(\dfrac{d}{dt}asin(2t+1) = \dfrac{1}{\sqrt{1-(2t+1)^{2}}}\cdot 2\)
See! That was very good. No need to fear. Just pay a little better attention.
\[2^t \frac{ 2 }{ \sqrt{1-(2t+1)}}+(\sin^{-1}(2t+1)(2^t* \log (2)) \]
Missed the "squared" in the square root. Otherwise, perfect.
Already showed you that. Pay better attention. \(\dfrac{2}{\sqrt{1 - (2t+1)^{2}}}\)
ok, thank you
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!
Thank you very much tkhunny. i will work hard and thank you for your advice.

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