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Mathematics 23 Online
OpenStudy (anonymous):

how to differentiate sin^2(2x+1)

OpenStudy (anonymous):

Let sin(2x+1) = u then dy/dx = 2u du/dx = 2sin(2x+1) d(sin(2x+1))/dx

OpenStudy (anonymous):

If you know the chain rule,then it's awesome.

OpenStudy (anonymous):

i know the chain rule but am getting 4sin(2x+1)

OpenStudy (anonymous):

answer given is 2sin(4x+2)

OpenStudy (anonymous):

it should be 4cos(2x+1) .. oh my god

OpenStudy (anonymous):

It is because : \[2\sin(x) \cos(x) = \sin(2x)\]

OpenStudy (anonymous):

Answer is right..

OpenStudy (anonymous):

thank you waterineyes i got my answer :D

OpenStudy (anonymous):

\[\frac{d}{dx}\sin^2(2x + 1) \implies 2\sin(2x + 1) \cdot \cos(2x + 1) \cdot 2\] \[\implies 2 \cdot 2\sin(2x + 1) \cos(2x + 1) \implies 2 \sin(2(2x + 1)) \implies 2 \sin(4x+2)\]

OpenStudy (anonymous):

You are welcome dear..

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