What is the tangent of f(x)=3e^xcos(x),\at x=0 ?

The tangent of f(x)=3e^xcos(x),\at x=0 is y=3x+3

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tangent of f(x)=3e^xcos(x),\at x=0

tangent of

f (x) = 3 e^{x} cos (x), at x = 0

y = 3 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = 3 e^{x} cos (x) : \frac{df ( x )}{d x} = 3 (e^{x} cos (x) - e^{x} sin (x))

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3

Find the line with slope m=

3

and passing through

(0, 3) : y = 3 x + 3

y = 3 x + 3

\frac{\partial}{\partial x} (x^{2} + \frac{y}{x})

\int \frac{4}{x ^{2} - 8 x - 9} d x

\frac{d}{d x} (\frac{11 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}})

\frac{x}{x + 1}

\frac{d y}{d x} sin (x) = y ln (y)

What is the tangent of f(x)=3e^xcos(x),\at x=0 ?
The tangent of f(x)=3e^xcos(x),\at x=0 is y=3x+3