Question 232975
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First apply the constant multiple rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (af)'\ =\ a\cdot f']


so


Given *[tex \LARGE f(x)\ =\ 2x^2] we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}(2x^2)\ =\ 2\frac{d}{dx}(x^2)] 


Next apply the power rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}(x^n)\ =\ nx^{n-1}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\frac{d}{dx}(x^2)\ =\ 2(2x^{2-1})\ =\ 4x]


And finally, to evaluate the first derivative at x = 1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ 4x]


therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(1)\ =\ 4(1)\ =\ 4]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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