let f be the function given by f(x)=2e^4x^2. for what value of x is the slope of the line tangent to the graph of f at (x,f(x)) equal to 3?
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$\displaystyle \begin{array}{l} \frac{{df}}{{dx}} = \frac{d}{{dx}}2e^{4x^2 } \\ u = 4x^2 \\ \frac{{df}}{{dx}} = \frac{{df}}{{du}}\frac{{du}}{{dx}} = 2e^u 8x = 16xe^{4x^2 } \\ \end{array} $
I understand how to get the derivative but now what?
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