What is the tangent of f(x)=2e^xcos(x),(0,2) ?

The tangent of f(x)=2e^xcos(x),(0,2) is y=2x+2

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tangent of f(x)=2e^xcos(x),(0,2)

tangent of

f (x) = 2 e^{x} cos (x), (0, 2)

y = 2 x + 2

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 2 e^{x} cos (x)

(0, 2) : m = 2

Find the line with slope m=

2

and passing through

(0, 2) : y = 2 x + 2

y = 2 x + 2

\frac{1}{3 x - 2}

n = 0 \sum \infty (\frac{- 1}{5})^{n}

\frac{\partial}{\partial x} (5 e^{6 x})

\int_{0}^{7} (7 - x)^{2} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{2}{3}

What is the tangent of f(x)=2e^xcos(x),(0,2) ?
The tangent of f(x)=2e^xcos(x),(0,2) is y=2x+2