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MIT 18.01 Single Variable Calculus (OCW)
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Would be the first derivative of this function with: F(X)=(1+2x+x^3)^1/4 ?
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\[f(x) = (1 + 2*x + x^3)^{1/4}\]
OK. \[f'(x) = (1/4)(1+2*x+x^3)^{-3/4} * ( 1 + 2 + 3x^2 )\]
the general concept is: g(f(x))'=g'(f(x)).f'(x)
oh, ok, I see. thanks
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f'(x)=[(1/4)(1+2∗x+x3)^−3/4].(2+3x^2)
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