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

Answer 1

Ezekiel Alexander

#(dy)/(dx)=2/(sqrt(1-(2x+1)^2))#

#y=sin^-1(2x+1)#

#=>2x+1=siny#

#:.2(dx)/dy=cosy#

#=>(dy)/dx=2/cosy#

using #cos^2y+sin^2y=1#

#(dy)/(dx)=2/(sqrt(1-sin^2y)#

substitue back for x

#(dy)/(dx)=2/(sqrt(1-(2x+1)^2))#

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Answer 2

Samantha Adkison

To find the derivative of the function y = arcsin(2x + 1), you can use the chain rule of differentiation.

The derivative of arcsin(u) with respect to u is 1 / sqrt(1 - u^2).

Using the chain rule, if y = arcsin(2x + 1), then u = 2x + 1.

So, the derivative of y with respect to x is: dy/dx = (1 / sqrt(1 - (2x + 1)^2)) * d(2x + 1)/dx.

Now, find d(2x + 1)/dx, which is simply 2.

Substitute this back into the expression: dy/dx = (1 / sqrt(1 - (2x + 1)^2)) * 2.

Therefore, the derivative of y = arcsin(2x + 1) with respect to x is: dy/dx = 2 / sqrt(1 - (2x + 1)^2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.