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How do you differentiate #sqrt(2x^3 - 3x- 4)#?

1 Answer

Dec 30, 2015

#(3(2x^2-1))/(2sqrt(2x^3-3x-4))#

Explanation:

This can be rewritten as #(2x^3-3x-4)^(1/2)#.

According to the chain rule, #d/dx(u^(1/2))=1/2u^(-1/2)*(du)/dx#.

Thus,

#d/dx(2x^3-3x-4)^(1/2)=1/2(2x^3-3x-4)^(-1/2)d/dx(2x^3-3x-4)#

#=>1/(2sqrt(2x^3-3x-4))*(6x^2-3)#

#=>(3(2x^2-1))/(2sqrt(2x^3-3x-4))#

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