What is the tangent of y=7e^{-6x},(0,7) ?

The tangent of y=7e^{-6x},(0,7) is y=-42x+7

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tangent of y=7e^{-6x},(0,7)

tangent of

y = 7 e^{- 6 x}, (0, 7)

y = - 42 x + 7

Solution steps

Find the slope of

y = 7 e^{- 6 x} : \frac{d y}{d x} = - 42 e^{- 6 x}

Compute the slope of

y = 7 e^{- 6 x}

(0, 7) : m = - 42

Find the line with slope m=

- 42

and passing through

(0, 7) : y = - 42 x + 7

y = - 42 x + 7

\frac{d y}{d x} = 9 x - \frac{y}{x}

y^{^{''}} - 2 y^{^{'}} = 4 t, y (0) = 2, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} (- x^{2} + y^{2} - 1)

\frac{d}{d x} ((6 x - x^{2})^{\frac{1}{2}})

\int t cos (t) sin (t) d t

What is the tangent of y=7e^{-6x},(0,7) ?
The tangent of y=7e^{-6x},(0,7) is y=-42x+7