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I've been given the function f(x)=x^(2/3)(x-5) and by using product rule came up with the first derivative f′(x)=5*(x-2)/3x^(1/3)
Now the following is how I tried to find the seconded derivative
f′′(x)=(3x^(1/3)*[5(x-2)]′-[3x^(1/3)]′*5(x-2))/[3x^(1/3)]^2
f′′(x)=(3x^(1/3)*5-x^(-2/3)*5(x-2))/9x^(2/3)
f′′(x)=(15x^(1/3)-5(x-2))/9x^(4/3)
From here I'm stuck as the answer I'm given is
f′′(x)=(10(x+1))/9x^4/3
I'm not sure how to continue from where I left off and get to this. i'm not exactly the most clever when it comes to math and could really use the help here.
Since we have f'(x) = g(x)/h(x), we know that f"(x) = [h(x)g'(x) - g(x)h'(x)]/h²(x).
Since g(x) = 5x-10, g'(x) = 5, and since h(x) = 3x^(1/3), h'(x) = 1/x^(2/3).
Putting these in gives [3x^(1/3)*5 - (5x-10)/x^(2/3)]/[9x^(2/3)].
Noting that 3*5 = 15 and multiplying the entire equation by
x^(2/3)/x^(2/3) gives [15x - (5x-10)]/[9x^(4/3)].
Distributing the minus sign on top gives (15x - 5x + 10)/[9x^(4/3)].
Combining like terms in the numerator gives (10x + 10)/[9x^(4/3)].
Factoring out a 10 in the numerator gives 10(x+1)/[9x^(4/3)].
That looks like the answer you gave me, so this is the math_guy, signing off.
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