How do you differentiate #y=xarcsinx+sqrt(1+x^2)#?

1 Answer
Mar 31, 2018

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

Explanation:

#dy/dx(xarcsinx+sqrt(1+x^2))=d/dxxarcsinx+d/dxsqrt(1+x^2)#

For #d/dxarcsinx,# recall that #d/dxarcsinx=1/sqrt(1-x^2)# and apply the product rule:

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

For #d/dxsqrt(1+x^2),# the Chain Rule will be needed:

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

Combining the above derivatives, we get

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