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Dodo1
 2 years ago
d/dt ( 2^(t)2sin^(1)(2t+1)
derivatives.
Dodo1
 2 years ago
d/dt ( 2^(t)2sin^(1)(2t+1) derivatives.

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tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Looks 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? \(\dfrac{d}{dt}2^{t+1}\cdot\sin^{1}(2t+1)\)

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0\[d/dt (2^t \sin ^{1}(2t+1)\]

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0product rule and chain rule?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, 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: \(\dfrac{d}{dt}2^{t} = 2^{t}\cdot\log(2)\)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1False. You had a clue a minute ago. Get it back.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0chain rule for each numbers?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Hint: \(\dfrac{d}{dt}\sin^{1}(t) = \dfrac{1}{\sqrt{1t^{2}}}\) That's about all you need.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Stop asking questions and start writing the derivative.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1\(\dfrac{d}{dt}f(t)\cdot g(t) = f(t)\cdot g'(t) + g(t)\cdot f'(t)\) Go!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1That's part of it. Keep going.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0product rules? but it does not make sense \[2^t \sin^{1} (2t+1) \] three parts. the product has two/ f(x)g(x)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1If there where three, you could just group them and make pairs. There are only two. \(2^{t}\) \(\sin^{1}(2t+1) = asin(2t+1)\)

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0\[2^t*(asin(2t+1)'+(asin(2t+1)(2^t)'\] ?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1There's that question again. Move forward. Do NOT stop to ask every time you flinch! Keep going!

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.1Right. 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?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0\[(2^t)(1/\sqrt(1t)*dx (2t+1)+asin(2t+1)(2^tlong(2)\]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1So Close!! \(\dfrac{d}{dt}asin(t) = \dfrac{1}{\sqrt{1t^{2}}}\) \(\dfrac{d}{dt}asin(2t+1) = \dfrac{1}{\sqrt{1(2t+1)^{2}}}\cdot 2\)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1See! That was very good. No need to fear. Just pay a little better attention.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0\[2^t \frac{ 2 }{ \sqrt{1(2t+1)}}+(\sin^{1}(2t+1)(2^t* \log (2)) \]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Missed the "squared" in the square root. Otherwise, perfect.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Already showed you that. Pay better attention. \(\dfrac{2}{\sqrt{1  (2t+1)^{2}}}\)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Now, 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!

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you very much tkhunny. i will work hard and thank you for your advice.
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