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

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Answer 1 · 2016-08-08T17:56:19

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

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

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

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

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

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

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

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