What is the tangent of y=e^{8x}cos(pix),(0,1) ?

The tangent of y=e^{8x}cos(pix),(0,1) is y=8x+1

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tangent of y=e^{8x}cos(pix),(0,1)

tangent of

y = e^{8 x} cos (π x), (0, 1)

y = 8 x + 1

Solution steps

Find the slope of

y = e^{8 x} cos (π x) : \frac{d y}{d x} = 8 e^{8 x} cos (π x) - π e^{8 x} sin (π 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 = 8

Find the line with slope m=

8

and passing through

(0, 1) : y = 8 x + 1

y = 8 x + 1

\frac{\partial}{\partial x} (sin (x) y)

\frac{\partial}{\partial y} (x ln (x y))

\int (sec (x))^{4} d x

\int 16 + 9 x^{2} d x

\frac{e ^{x} d x}{- ( e ^{x} + 1 ) ^{3}} = \frac{e ^{y} d y}{( e ^{y} + 1 ) ^{2}}

What is the tangent of y=e^{8x}cos(pix),(0,1) ?
The tangent of y=e^{8x}cos(pix),(0,1) is y=8x+1