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

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Answer 1 · 2018-03-31T20:20:28

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

#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)#