What is the tangent of f(x)=9e^5x+e^{-x^2},\at x=1 ?

The tangent of f(x)=9e^5x+e^{-x^2},\at x=1 is y=(9e^5-2/e)x+3/e

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tangent of f(x)=9e^5x+e^{-x^2},\at x=1

tangent of

f (x) = 9 e^{5} x + e^{- x^{2}}, at x = 1

y = (9 e^{5} - \frac{2}{e}) x + \frac{3}{e}

Solution steps

Find the tangent point:

(1, 9 e^{5} + \frac{1}{e})

Find the slope of

f (x) = 9 e^{5} x + e^{- x^{2}} : \frac{df ( x )}{d x} = 9 e^{5} - 2 e^{- x^{2}} x

Compute the slope of

f (x) = 9 e^{5} x + e^{- x^{2}}

x = 1 : m = 9 e^{5} - \frac{2}{e}

Find the line with slope m=

9 e^{5} - \frac{2}{e}

and passing through

(1, 9 e^{5} + \frac{1}{e}) : y = (9 e^{5} - \frac{2}{e}) x + \frac{3}{e}

y = (9 e^{5} - \frac{2}{e}) x + \frac{3}{e}

\frac{d y}{d x} = 6 x^{2} y^{2}

x \to 2 - lim (4 - x^{2})

f (r) = \frac{5}{r ^{3}}

\frac{d}{d x} (2 x y + y)

\frac{\partial}{\partial t} (s t)

What is the tangent of f(x)=9e^5x+e^{-x^2},\at x=1 ?
The tangent of f(x)=9e^5x+e^{-x^2},\at x=1 is y=(9e^5-2/e)x+3/e