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

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

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

tangent of

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

y = 2 x + 2

Solution steps

Find the tangent point:

(0, 2)

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)

x = 0 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 2

\int 0.67 sin (100 π x + \frac{π}{3}) d x

\int ((1 + \frac{1}{x})^{3} (\frac{1}{x ^{2}})) d x

8 + x^{2}

f (t) = - cos (t)

ln (5 - x)

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