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

Question

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

Aurora Alvarado · Accepted Answer

It is #f'(x)=2/(sqrt(1-(2x+1)^2))#

Let # u(x) = 2x + 1#, function f may be considered as the composition #f(x) = arcsin(u(x))#. Hence we use the chain rule, #f '(x) = ((du)/dx) (d(arcsin(u)))/(du)#, to differentiate function f as follows

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

Charles Anctil · Answer

undefined 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}} $.

Evelyn Agard · Answer

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

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

Using these steps:

1. Inner function: u(x) = 2x + 1.
2. Derivative of inner function: u'(x) = 2.
3. Derivative of arcsin function: d(arcsin(u))/du = 1 / sqrt(1 - u^2).
4. Apply chain rule: f'(x) = u'(x) * d(arcsin(u))/du = 2 * (1 / sqrt(1 - (2x + 1)^2)).
5. 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).