What is the tangent of f(x)=e^{9x}cos(pix),\at x=0 ?

The tangent of f(x)=e^{9x}cos(pix),\at x=0 is y=9x+1

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tangent of f(x)=e^{9x}cos(pix),\at x=0

tangent of

f (x) = e^{9 x} cos (π x), at x = 0

y = 9 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

f (x) = e^{9 x} cos (π x)

x = 0 : m = 9

Find the line with slope m=

9

and passing through

(0, 1) : y = 9 x + 1

y = 9 x + 1

9 x^{8} e^{x} + e^{x} x^{9}

\int (x^{4} - e^{- 2 x}) d x

\int cos (t) tan (t) d t

\frac{d}{d x} (x^{4} e^{4 x})

y^{^{'}} = y (x y^{4} - 1)

What is the tangent of f(x)=e^{9x}cos(pix),\at x=0 ?
The tangent of f(x)=e^{9x}cos(pix),\at x=0 is y=9x+1