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$g(x) = e^{(x^2-x)}$ $g'(x) = e^{(x^2-x)}(2x-1)$ But if I use implicit differentiation why isn't this the equivalent? $y = e^{(x^2-x)}$ $ln(y) = (x^2-x)ln(e)$ $\frac{1}{y} * y' = (x^2-x)(\frac{1}{e})+ln(e)(2x-1)$ $\frac{y'}{y} = (\frac{(x^2-x)}{e})+ln(e)(2x-1)$ $y' = ((\frac{(x^2-x)}{e})+ln(e)(2x-1))y$ $y' = ((\frac{(x^2-x)}{e})+ln(e)(2x-1))e^{(x^2-x)}$
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Tue, 28 Mar 2023 01:57 GMT