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How do you differentiate #root3(x)+root4(x)#?

1 Answer

Aug 8, 2016

#=1/(3x^(2/3))+1/(4x^(3/4))#.

Explanation:

We know that, #d/dx(x^n)=n*x^(n-1)#, and,

#d/dx(u+v)=(du)/dx+(dv)/dx#

Therefore, #d/dx(root3x+root4x)=d/dxroot3x+d/dxroot4x#

#=d/dxx^(1/3)+d/dxx^(1/4)#

#=1/3*x^(1/3-1)+1/4*x^(1/4-1)#

#=1/(3x^(2/3))+1/(4x^(3/4))#.

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