What is the tangent of y=7e^{1-x^2},\at ?

The tangent of y=7e^{1-x^2},\at is y=-14e^{1-a_{0}^2}a_{0}x+7e^{1-a_{0}^2}+14a_{0}^2e^{1-a_{0}^2}

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tangent of y=7e^{1-x^2},\at

tangent of

y = 7 e^{1 - x^{2}}, at

y = - 14 e^{1 - a_{0}^{2}} a_{0} x + 7 e^{1 - a_{0}^{2}} + 14 a_{0}^{2} e^{1 - a_{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 7 e^{1 - a_{0}^{2}})

Find the slope of

y = 7 e^{1 - x^{2}} : \frac{d y}{d x} = - 14 e^{1 - x^{2}} x

Compute the slope of

y = 7 e^{1 - x^{2}}

x = a_{0} : m = - 14 e^{1 - a_{0}^{2}} a_{0}

Find the line with slope m=

- 14 e^{1 - a_{0}^{2}} a_{0}

and passing through

(a_{0}, 7 e^{1 - a_{0}^{2}}) : y = - 14 e^{1 - a_{0}^{2}} a_{0} x + 7 e^{1 - a_{0}^{2}} + 14 a_{0}^{2} e^{1 - a_{0}^{2}}

y = - 14 e^{1 - a_{0}^{2}} a_{0} x + 7 e^{1 - a_{0}^{2}} + 14 a_{0}^{2} e^{1 - a_{0}^{2}}

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

x \frac{d y}{d x} + (3 x + 1) y = e^{3 x}

(6 y + 2 t - 7) d t + (8 y + 6 t - 1) d y = 0

\int \frac{x}{3 - 2 x - x ^{2}} d x

y^{^{'}} = t y^{2}

What is the tangent of y=7e^{1-x^2},\at ?
The tangent of y=7e^{1-x^2},\at is y=-14e^{1-a_{0}^2}a_{0}x+7e^{1-a_{0}^2}+14a_{0}^2e^{1-a_{0}^2}