# Let f (x) = e^xcos x . Find the gradient of the normal to the curve of f at x = π

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### 1 Answer

`f(x)=e^xcos(x)`

differentiate f(x) with respect to x

`f'(x)=e^x cos(x)-e^xsin(x)`

`f'(x=pi)=e^(pi)cos(pi)-e^(pi) sin(pi)`

`=e^(pi)(-1)`

`=-e^(pi)`

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`

`-e^(pi)xxm=-1`

`m=e^(-pi)`

Thus gradient of the normal to the curve f at x=pi is equal to `e^(-pi)`