Asked by iain
DERIVATIVE of:
g(x) = Sq.root of x^4-8x^3+9
thanks
g(x) = Sq.root of x^4-8x^3+9
thanks
Answers
Answered by
drwls
Let u(x) = x^4-8x^3+9
g(x) = [u(x)]^1/2
Use the chain rule.
dg(x)/dx = dg/du * du/dx
= [1/2u(x)^(1/2)]*(3x^3 -24x^2)
= (3/2)*x^2(x-8)/[x^4-8x^3+9]^1/2
g(x) = [u(x)]^1/2
Use the chain rule.
dg(x)/dx = dg/du * du/dx
= [1/2u(x)^(1/2)]*(3x^3 -24x^2)
= (3/2)*x^2(x-8)/[x^4-8x^3+9]^1/2
Answered by
Steve
I think we ran aground at:
dg(x)/dx = dg/du * du/dx
= [1/2u(x)^(1/2)]*(3x^3 -24x^2)
= (3/2)*x^2(x-8)/[x^4-8x^3+9]^1/2
where it should be
= [1/2 u^-1/2](4x^3 - 24x^2)
= 2x^2 (x-6)/(x^4-8x^3+9)^1/2
dg(x)/dx = dg/du * du/dx
= [1/2u(x)^(1/2)]*(3x^3 -24x^2)
= (3/2)*x^2(x-8)/[x^4-8x^3+9]^1/2
where it should be
= [1/2 u^-1/2](4x^3 - 24x^2)
= 2x^2 (x-6)/(x^4-8x^3+9)^1/2
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