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

The tangent of f(x)=e^{x^2},\at x= 1/2 is y=e^{1/4}x+(e^{1/4})/2

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tangent of f(x)=e^{x^2},\at x= 1/2

tangent of

f (x) = e^{x^{2}}, at x = \frac{1}{2}

y = e^{\frac{1}{4}} x + \frac{e ^{\frac{1}{4}}}{2}

Solution steps

Find the tangent point:

(\frac{1}{2}, e^{\frac{1}{4}})

Find the slope of

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

Compute the slope of

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

x = \frac{1}{2} : m = e^{\frac{1}{4}}

Find the line with slope m=

e^{\frac{1}{4}}

and passing through

(\frac{1}{2}, e^{\frac{1}{4}}) : y = e^{\frac{1}{4}} x + \frac{e ^{\frac{1}{4}}}{2}

y = e^{\frac{1}{4}} x + \frac{e ^{\frac{1}{4}}}{2}

\int \frac{1}{p} d p

\int \frac{4 x ^{2} - x + 25}{x ^{3} + 25 x} d x

1 + \frac{1}{x}

\frac{d}{d x} (\frac{9}{x ^{2}} - \frac{3}{x})

7 y ln (x) - x \frac{d y}{d x} = 0

What is the tangent of f(x)=e^{x^2},\at x= 1/2 ?
The tangent of f(x)=e^{x^2},\at x= 1/2 is y=e^{1/4}x+(e^{1/4})/2