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

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

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tangent of y=9e^xcos(x),(0,9)

tangent of

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

y = 9 x + 9

Solution steps

Find the slope of

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

Compute the slope of

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

(0, 9) : m = 9

Find the line with slope m=

9

and passing through

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

y = 9 x + 9

\frac{d}{d x} ((x + sin (x))^{5})

3 tan (x), at x = \frac{π}{3}

\frac{d}{d θ} (2 sin (θ) + 2 cos (θ))

(9.5) (21.1)

\frac{d ^{2}}{d x ^{2}} (5 (x^{2} + 3 x)^{3})

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