What is the tangent of f(x)=sqrt(1-2x) ?

The tangent of f(x)=sqrt(1-2x) is y=-1/(sqrt(1-2a_{0))}x+sqrt(1-2a_{0)}+(a_{0})/(sqrt(1-2a_{0))}

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tangent of f(x)=sqrt(1-2x)

tangent of

f (x) = 1 - 2 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 1 - 2 a_{0})

Find the slope of

f (x) = 1 - 2 x : \frac{df ( x )}{d x} = - \frac{1}{1 - 2 x}

Compute the slope of

f (x) = 1 - 2 x

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

Find the line with slope m=

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

and passing through

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

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

\frac{d}{d x} (e^{- 3 x} sin (3 x))

y = x^{2} + x + 1

\frac{\partial}{\partial y} (x + \frac{1}{2} ln (x^{2} + y^{2}))

\frac{1}{y}

x \to 1 lim (\frac{e ^{x}}{2})

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