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

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

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tangent of y=e^{9x}cos(pix),(0,1)

tangent of

y = e^{9 x} cos (π x), (0, 1)

y = 9 x + 1

Solution steps

Find the slope of

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

Compute the slope of

y = e^{9 x} cos (π x)

(0, 1) : m = 9

Find the line with slope m=

9

and passing through

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

y = 9 x + 1

ln (2 - x)

3 x^{3}

y^{^{'}} = x ln (x)

\frac{d}{d x} (\frac{8 x}{5})

\frac{d}{d x} ((3 x^{6} + 2 x^{3})^{4})

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