Question 681860
find derivative of (t^2/3)/(3t-5)
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Is it {{{t^2/3}}} ?
Or {{{t^(2/3)}}} ?
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I think the latter.
{{{f(t) = (t^(2/3))*(3t-5)^(-1)}}}
f'(t) = {{{(2/3)*t^(-1/3)*(3t-5)^(-1) + t^(2/3)*-1*(3t-5)^-2*3}}}
= {{{(2/3)*t^(-1/3)*(3t-5)^(-1) - 3t^(2/3)*(3t-5)^-2}}}
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= {{{(-3t-10)/(3*root(3,t)*(3t-5)^2)}}}