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differentiate f(x) with respect to x
product of slope of tangent and slope of normal at a is -1.
But slope of tangent at x=pi on the curve f(x)= f'(pi)=-e^(pi)
Let slope of the normal be m
`f(pi) xx m=-1`
Thus gradient of the normal to the curve f at x=pi is equal to `e^(-pi)`
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