What is the tangent of sqrt(1-4x) ?

The tangent of sqrt(1-4x) is y=-2/(sqrt(1-4a_{0))}x+sqrt(1-4a_{0)}+(2a_{0})/(sqrt(1-4a_{0))}

tangent of sqrt(1-4x)

tangent of

1 - 4 x

y = - \frac{2}{1 - 4 a _{0}} x + 1 - 4 a_{0} + \frac{2 a _{0}}{1 - 4 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 1 - 4 a_{0})

Find the slope of

y = 1 - 4 x : \frac{d y}{d x} = - \frac{2}{1 - 4 x}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{2}{1 - 4 a _{0}}

Find the line with slope m=

- \frac{2}{1 - 4 a _{0}}

and passing through

(a_{0}, 1 - 4 a_{0}) : y = - \frac{2}{1 - 4 a _{0}} x + 1 - 4 a_{0} + \frac{2 a _{0}}{1 - 4 a _{0}}

y = - \frac{2}{1 - 4 a _{0}} x + 1 - 4 a_{0} + \frac{2 a _{0}}{1 - 4 a _{0}}

n = 1 \sum \infty \frac{( 1 )}{n ^{2}}

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

\frac{d}{d x} (\frac{e ^{x}}{x})

x \to 0 + lim (arcsin (2 x - 1))

im pl i c i t \frac{d y}{d x}, 3 x^{3} y = 6

What is the tangent of sqrt(1-4x) ?
The tangent of sqrt(1-4x) is y=-2/(sqrt(1-4a_{0))}x+sqrt(1-4a_{0)}+(2a_{0})/(sqrt(1-4a_{0))}