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

The tangent of f(x)=e^xcos(x),\at x=0 is y=x+1

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

tangent of

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

y = x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = e^{x} cos (x) : \frac{df ( x )}{d x} = 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 = 1

Find the line with slope m=

1

and passing through

(0, 1) : y = x + 1

y = x + 1

\int 4 csc (x) cot (x) d x

y = 8^{x} \cdot lo g_{7} (x)

\int e^{7 x} sin (4 x) d x

f (x) = 1 + e^{2 x}

x + x^{2} + 1

What is the tangent of f(x)=e^xcos(x),\at x=0 ?
The tangent of f(x)=e^xcos(x),\at x=0 is y=x+1