What is the tangent of f(x)=7e^xcos(x),\at x=0 ?

The tangent of f(x)=7e^xcos(x),\at x=0 is y=7x+7

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tangent of f(x)=7e^xcos(x),\at x=0

tangent of

f (x) = 7 e^{x} cos (x), at x = 0

y = 7 x + 7

Solution steps

Find the tangent point:

(0, 7)

Find the slope of

f (x) = 7 e^{x} cos (x) : \frac{df ( x )}{d x} = 7 (e^{x} cos (x) - e^{x} sin (x))

Compute the slope of

f (x) = 7 e^{x} cos (x)

x = 0 : m = 7

Find the line with slope m=

7

and passing through

(0, 7) : y = 7 x + 7

y = 7 x + 7

x \to 0 lim ((\frac{x - 4}{x + 1})^{x - 2})

\frac{d}{d x} (\frac{x x}{x ^{2}})

y^{^{'}} + 3 y = e^{- 4 t}, y (0) = 5

\frac{d}{d x} (\frac{9}{e ^{x} - e ^{- x}})

x^{2} + 3 y \frac{d y}{d x} = 0

What is the tangent of f(x)=7e^xcos(x),\at x=0 ?
The tangent of f(x)=7e^xcos(x),\at x=0 is y=7x+7