# How do you differentiate # f(x) =arcsin(2x + 1) #?

It is

#f'(x)=((2x+1)/dx)*(1/(sqrt(1-u^2))) => f'(x)=2/(sqrt(1-(2x+1)^2))#

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To differentiate the function ( f(x) = \arcsin(2x + 1) ), you use the chain rule, which states that if ( u ) is a function of ( x ), and ( y ) is a function of ( u ), then the derivative of ( y ) with respect to ( x ) is ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ). Applying this rule:

Let ( u = 2x + 1 ), and ( y = \arcsin(u) ). ( \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}} ) (derivative of ( \arcsin(u) )) ( \frac{du}{dx} = 2 ) (derivative of ( 2x + 1 ))

Applying the chain rule: ( \frac{dy}{dx} = \frac{1}{\sqrt{1 - (2x + 1)^2}} \cdot 2 )

So, the derivative of ( f(x) = \arcsin(2x + 1) ) is ( \frac{2}{\sqrt{1 - (2x + 1)^2}} ).

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To differentiate f(x) = arcsin(2x + 1), we use the chain rule:

- Identify the inner function u(x) = 2x + 1.
- Find the derivative of the inner function: u'(x) = 2.
- Apply the derivative of arcsin function: d(arcsin(u))/du = 1 / sqrt(1 - u^2).
- Multiply the results of steps 2 and 3 to get the derivative of the composition.
- Substitute the inner function u(x) back into the result.

Using these steps:

- Inner function: u(x) = 2x + 1.
- Derivative of inner function: u'(x) = 2.
- Derivative of arcsin function: d(arcsin(u))/du = 1 / sqrt(1 - u^2).
- Apply chain rule: f'(x) = u'(x) * d(arcsin(u))/du = 2 * (1 / sqrt(1 - (2x + 1)^2)).
- Substitute back the inner function: f'(x) = 2 / sqrt(1 - (2x + 1)^2).

Therefore, the derivative of f(x) = arcsin(2x + 1) is f'(x) = 2 / sqrt(1 - (2x + 1)^2).

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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.

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.

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.

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