What is the tangent of f(x)=-2xe^{x^2},\at x=2 ?

The tangent of f(x)=-2xe^{x^2},\at x=2 is y=-18e^4x+32e^4

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tangent of f(x)=-2xe^{x^2},\at x=2

tangent of

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

y = - 18 e^{4} x + 32 e^{4}

Solution steps

Find the tangent point:

(2, - 4 e^{4})

Find the slope of

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

Compute the slope of

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

x = 2 : m = - 18 e^{4}

Find the line with slope m=

- 18 e^{4}

and passing through

(2, - 4 e^{4}) : y = - 18 e^{4} x + 32 e^{4}

y = - 18 e^{4} x + 32 e^{4}

n = 4 \sum \infty \frac{x ^{n}}{n ^{5}}

x - 1 (x^{2} + 1)

x \to 0 + lim (6 - \frac{1}{x ^{3}})

t^{2} y^{^{''}} - 8 t y^{^{'}} + 8 y = 0

\int 6 x ln (x) d x

What is the tangent of f(x)=-2xe^{x^2},\at x=2 ?
The tangent of f(x)=-2xe^{x^2},\at x=2 is y=-18e^4x+32e^4