DERIVATIVE of:
g(x) = Sq.root of x^4-8x^3+9
thanks
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
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
To find the derivative of the function g(x) = √(x^4 - 8x^3 + 9), you can use the chain rule and power rule for differentiation.
First, let's rewrite the function as g(x) = (x^4 - 8x^3 + 9)^(1/2).
Now applying the chain rule, the derivative of g(x) with respect to x is given by:
g'(x) = (1/2)(x^4 - 8x^3 + 9)^(-1/2) * (4x^3 - 24x^2).
To simplify further, you can rewrite the expression as:
g'(x) = (4x^3 - 24x^2) / (2√(x^4 - 8x^3 + 9)).
So, the derivative of g(x) is g'(x) = (4x^3 - 24x^2) / (2√(x^4 - 8x^3 + 9)).
That's how you find the derivative of the given function using the chain rule and power rule.