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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

\[d/dt (2^t \sin ^{-1}(2t+1)\]

product rule and chain rule?

mmm i have no clue.

False. You had a clue a minute ago. Get it back.

ok,

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.

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!

\[(2^t)(1/\sqrt(1-t)*dx (2t+1)+asin(2t+1)(2^tlong(2)\]

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.

sqrt(2t+1)?

Already showed you that. Pay better attention.
\(\dfrac{2}{\sqrt{1 - (2t+1)^{2}}}\)

ok, thank you

Thank you very much tkhunny.
i will work hard and thank you for your advice.