What is the tangent of x^2=e^{(x-y)} ?

The tangent of x^2=e^{(x-y)} is y=(1-2/(a_{0)})x-2ln(a_{0})+2

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tangent of x^2=e^{(x-y)}

tangent of

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

y = (1 - \frac{2}{a _{0}}) x - 2 ln (a_{0}) + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} - 2 ln (a_{0}))

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 1 - \frac{2}{a _{0}}

Find the line with slope m=

1 - \frac{2}{a _{0}}

and passing through

(a_{0}, a_{0} - 2 ln (a_{0})) : y = (1 - \frac{2}{a _{0}}) x - 2 ln (a_{0}) + 2

y = (1 - \frac{2}{a _{0}}) x - 2 ln (a_{0}) + 2

f (t) = \frac{1}{t}

\frac{d}{d t} (6 t)

v^{\frac{3}{2}} (v + c e^{v})

\frac{d}{d x} (\frac{5 x - 2}{4 x + 3})

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

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