How do you find the derivative of #(arcsin(x))^(2)#?

Question

Charles Arocha · Accepted Answer

#(2arcsinx)/sqrt(1-x^2)#

We have #d/dxarcsin^2x#

According to the chain rule, #(df)/dx=(df)/(du)*(du)/dx#, where #u# is a function within #f#.

Here, #f=u^2# where #u=arcsinx#

So we have #d/(du)u^2*d/dxarcsinx#

#2u*d/dxarcsin x#

To find the derivative of #arcsinx#, take #arcsinx=sin^-1x#. We have:

#y=sin^-1x#, and so:

#siny=x#

Taking the derivative of both sides, we have:

#(cosy)*y'=1#

#y'=1/cosy#

We need to write #cosy# in terms of #siny#, in order to change the variable to #x#.

Remember that #sin^2y+cos^2y=1#. So:

#cos^2y=1-sin^2y#

#cosy=sqrt(1-sin^2y)#

We can take the positive square root, as #sin^-1x# is normally given on the interval #-pi/2,pi/2# and as the cosine function is positive in that interval.

Inputting that into our earlier function:

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

#y'=1/sqrt(1-x^2)#

So as #d/dxarcsinx=1/sqrt(1-x^2)#, we can write the derivative of #arcsin^2x# as:

#2u*1/sqrt(1-x^2)#

But remember that #u=arcsinx#, so we have:

#(2arcsinx)/sqrt(1-x^2)#

Done.

Emilia Aguilar · Answer

undefined To find the derivative of $(\arcsin(x))^2$, we can use the chain rule. Let $f(x) = \arcsin(x)$. Then, $f'(x) = \frac{1}{\sqrt{1-x^2}}$.

Now, applying the chain rule, we have:

$$\frac{d}{dx} (\arcsin(x))^2 = 2 \arcsin(x) \cdot \frac{1}{\sqrt{1-x^2}}$$