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

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Answer 1 · 2015-12-30T04:27:54

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

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,

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

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

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

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