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

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

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

tangent of

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

y = 9 x + 9

Solution steps

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)

(0, 9) : m = 9

Find the line with slope m=

9

and passing through

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

y = 9 x + 9

y^{^{''}} + y = x^{3}

\int_{0}^{y} y e^{- y} d y

\frac{d y}{d t} y = (2 t - 1) (3 t - 4)^{- 1}

lo g_{5} (2 x)

\int_{- \frac{π}{4}}^{\frac{π}{4}} tan (x) d x

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