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How do you find the derivative of y=arcsin(2x+1)y=arcsin(2x+1)?

User Gbrennon
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2 Answers

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y=arcsin(2x+1)\\\\siny=2x+1\ (*)\\\\2x=siny-1\\\\x=(1)/(2)siny-(1)/(2)\\\\(dx)/(dy)=(1)/(2)cosy\Rightarrow(dy)/(dx)=(2)/(cosy)=(2)/(√(1-sin^2y))\\\\substitute\ (*)\\\\=(2)/(√(1-(2x+1)^2))=(2)/(√(1-4x^2-4x-1))=(2)/(√(-4x^2-4x))=(2)/(√(4(-x^2-x)))\\\\=(2)/(\sqrt4\cdot√(-x^2-x))=(2)/(2√(-x(x+1)))\\\center\boxed{=(1)/(√(-x(x+1)))}
User Rusi Nova
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3 votes
Answer is in the attachment below.
How do you find the derivative of y=arcsin(2x+1)y=arcsin(2x+1)?-example-1
User Mike Mathew
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