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Find the tangent line to the curve y= e^(2-x^2) when x = 1.
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let 2-x^2=u -> y'=e^udu -> du = -2x -> y'=-2xe^(2-x^2) slope of the tangent line is -2e^2 now find a few points on the line and write the line's equation.
y = -2ex + 3e the slope is actually -2e, right? becayse e^(2-1^2)=e
Yes, the slope is \(-2e\).
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