What is the tangent of f(x)=9e^xcos(x),\at x=0 ?

The tangent of f(x)=9e^xcos(x),\at x=0 is y=9x+9

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tangent of f(x)=9e^xcos(x),\at x=0

tangent of

f (x) = 9 e^{x} cos (x), at x = 0

y = 9 x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

f (x) = 9 e^{x} cos (x) : \frac{df ( x )}{d x} = 9 (e^{x} cos (x) - e^{x} sin (x))

Compute the slope of

f (x) = 9 e^{x} cos (x)

x = 0 : m = 9

Find the line with slope m=

9

and passing through

(0, 9) : y = 9 x + 9

y = 9 x + 9

\frac{\partial}{\partial v} (4 u cos (v))

\frac{\partial}{\partial x} (\frac{1}{1 + e ^{- x}})

\int \frac{sin ( 3 x )}{3} d x

\int \frac{- x}{1 + x} d x

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

What is the tangent of f(x)=9e^xcos(x),\at x=0 ?
The tangent of f(x)=9e^xcos(x),\at x=0 is y=9x+9