How do I find the derivative of # y=xsin^(-1)x + sqrt(1-x^2)#?

Question

How do I find the derivative of #  y=xsin^(-1)x + sqrt(1-x^2)#?

Aurora Alvarado · Accepted Answer

#y^' = arcsinx#

You can differentiate this function by using the product rule and the chain rule, provided that you know that

#d/dx(arcsinx) = 1/(sqrt(1-x^2)#

So, your function can be written like this

#y = x * arcsinx + sqrt(1-x^2)#

You will use the product rule to differentiate #x * arcsinx#, and the chain rule to differentiate #sqrt(u)#, with #u = 1-x^2#. This will get you

#d/dx(x * arcsinx) = [d/dx(x)] * arcsinx + x * d/dx(arcsinx)#

#d/dx(x * arcsinx) = 1 * arcsinx + x * 1/(sqrt(1-x^2)#

and

#d/dx(sqrt(u)) = d/(du)sqrt(u) * d/dx(u)#

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

#d/dx(sqrt(1-x^2)) = 1/color(red)(cancel(color(black)(2))) * 1/sqrt(1-x^2) * (-color(red)(cancel(color(black)(2)))x)#

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

The derivative of #y# will thus be

#d/dx(y) = d/dx(x * arcsinx) + d/dx(sqrt(1-x^2))#

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

#y^' = arcsinx + color(red)(cancel(color(black)(x/sqrt(1-x^2)))) -color(red)(cancel(color(black)(x/sqrt(1-x^2))))#

#y^' = color(green)(arcsinx)#

Isabella Ackerman · Answer

undefined To find the derivative of $y = x\sin^{-1}x + \sqrt{1-x^2}$, we can use the sum rule and the chain rule. The derivative is $\sin^{-1}x + \frac{x}{\sqrt{1-x^2}}$.