What is the derivative of #f(x)=arcsin sqrt sinx#?

Question

Ezekiel Alexander · Accepted Answer

One can derive the derivative for #arcsinx# with implicit differentiation if it is not easy to remember it. #y = arcsinx#
#siny = x#
#cosy((dy)/(dx)) = 1#
#(dy)/(dx) = 1/(cosy) = 1/(sqrt(1-sin^2y)) = 1/(sqrt(1-x^2))# since #sin^2x + cos^2x = 1#. Thus, take this further with the Chain Rule. #d/(dx)[arcsinsqrt(sinx)] = 1/(sqrt(1-(sqrtsinx)^2)) * 1/((2sqrtsinx)) * cosx# #= color(blue)(cosx/(2sqrtsinxsqrt(1-sinx)))#

Emma Aldridge · Answer

undefined The derivative of $ f(x) = \arcsin(\sqrt{\sin(x)}) $ can be found using the chain rule:

1. Identify the outer function $ u $ and the inner function $ v $.
   In this case, the outer function is $ u = \arcsin(v) $ and the inner function is $ v = \sqrt{\sin(x)} $.

2. Find the derivative of the inner function $ v $ with respect to $ x $, denoted as $ \frac{dv}{dx} $.
   $ \frac{dv}{dx} = \frac{1}{2\sqrt{\sin(x)}}\cdot \cos(x) $.

3. Find the derivative of the outer function $ u $ with respect to $ v $, denoted as $ \frac{du}{dv} $.
   $ \frac{du}{dv} = \frac{1}{\sqrt{1-v^2}} $.

4. Apply the chain rule formula $ \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} $.
   Substitute the derivatives found in steps 2 and 3 into the chain rule formula.
   $ \frac{dy}{dx} = \frac{1}{\sqrt{1-\sin(x)}} \cdot \frac{1}{2\sqrt{\sin(x)}}\cdot \cos(x) $.

5. Simplify the expression.
   $ \frac{dy}{dx} = \frac{\cos(x)}{2\sqrt{\sin(x)(1-\sin(x))}} $.

Therefore, the derivative of $ f(x) = \arcsin(\sqrt{\sin(x)}) $ with respect to $ x $ is $ \frac{\cos(x)}{2\sqrt{\sin(x)(1-\sin(x))}} $.