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mezeva1999
@mezeva1999
July 2022
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Найдите тангенс угла наклона касательной к графику функции y=h(x) в точке x0:
h(x)=(1/e)^x, x0=2
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sweetburning
Y'=-e^(-x)
y'(2)=-e^(-2)=-1/e^2
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mezeva1999
Спасибо большое
mezeva1999
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Answers & Comments
y'(2)=-e^(-2)=-1/e^2