What is the tangent of y=8e^xcos^2(x),\at x= pi/4 ?

The tangent of y=8e^xcos^2(x),\at x= pi/4 is y=-4e^{pi/4}x+4e^{pi/4}+pie^{pi/4}

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tangent of y=8e^xcos^2(x),\at x= pi/4

tangent of

y = 8 e^{x} cos^{2} (x), at x = \frac{π}{4}

y = - 4 e^{\frac{π}{4}} x + 4 e^{\frac{π}{4}} + π e^{\frac{π}{4}}

Solution steps

Find the tangent point:

(\frac{π}{4}, 4 e^{\frac{π}{4}})

Find the slope of

y = 8 e^{x} cos^{2} (x) : \frac{d y}{d x} = 8 (e^{x} cos^{2} (x) - e^{x} sin (2 x))

Compute the slope of

y = 8 e^{x} cos^{2} (x)

x = \frac{π}{4} : m = - 4 e^{\frac{π}{4}}

Find the line with slope m=

- 4 e^{\frac{π}{4}}

and passing through

(\frac{π}{4}, 4 e^{\frac{π}{4}}) : y = - 4 e^{\frac{π}{4}} x + 4 e^{\frac{π}{4}} + π e^{\frac{π}{4}}

y = - 4 e^{\frac{π}{4}} x + 4 e^{\frac{π}{4}} + π e^{\frac{π}{4}}

\frac{\partial}{\partial x} (- 2 cos (- 3 x + 7))

\frac{d}{d t} (2 t^{- 1})

f (x) = 2 e x

(1 + \frac{1}{x}) \frac{d y}{d x} + y = 0

x \to 0 - lim (\frac{tan ( x )}{x})

What is the tangent of y=8e^xcos^2(x),\at x= pi/4 ?
The tangent of y=8e^xcos^2(x),\at x= pi/4 is y=-4e^{pi/4}x+4e^{pi/4}+pie^{pi/4}