What is the tangent of f(x)=e^{-2x},(0,1) ?

The tangent of f(x)=e^{-2x},(0,1) is y=-2x+1

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tangent of f(x)=e^{-2x},(0,1)

tangent of

f (x) = e^{- 2 x}, (0, 1)

y = - 2 x + 1

Solution steps

Find the slope of

f (x) = e^{- 2 x} : \frac{df ( x )}{d x} = - 2 e^{- 2 x}

Compute the slope of

f (x) = e^{- 2 x}

(0, 1) : m = - 2

Find the line with slope m=

- 2

and passing through

(0, 1) : y = - 2 x + 1

y = - 2 x + 1

y^{^{''}} - 2 y^{^{'}} + 10 y = e^{- x}

e^{- x + 1} - e^{- x + 1} \cdot x

\frac{\partial}{\partial x} (y \cdot cos (x - y))

7 x - 4 x^{2}

What is the tangent of f(x)=e^{-2x},(0,1) ?
The tangent of f(x)=e^{-2x},(0,1) is y=-2x+1