What is the tangent of f(x)=(1+2x)/(1-2x)enel,(1,-3) ?

The tangent of f(x)=(1+2x)/(1-2x)enel,(1,-3) is

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tangent of f(x)=(1+2x)/(1-2x)enel,(1,-3)

tangent of

f (x) = \frac{1 + 2 x}{1 - 2 x} e n e l, (1, - 3)

y = - \frac{2 ( 3 e ^{4} l ^{2} n ^{2} + 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n - 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3} x - 3 + \frac{2 ( 3 e ^{4} l ^{2} n ^{2} + 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n - 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3} or y = - \frac{2 ( 3 e ^{4} l ^{2} n ^{2} - 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n + 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n - 6 e l n ( e ^{2} l n - 1 ) + 3} x - 3 + \frac{2 ( 3 e ^{4} l ^{2} n ^{2} - 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n + 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n - 6 e l n ( e ^{2} l n - 1 ) + 3}

Solution steps

f (x) = \frac{1 + 2 x}{1 - 2 x} e n e l : f (x) = \frac{e ^{2} l n ( 1 + 2 x )}{1 - 2 x}

Slope of the tangent line at

(1, - 3) : m = \frac{e ^{2} l n ( 1 + 2 x ) + 3 ( 1 - 2 x )}{( 1 - 2 x ) ( x - 1 )}

Find the slope of

f (x) = \frac{e ^{2} l n ( 1 + 2 x )}{1 - 2 x} : m = \frac{4 e ^{2} l n}{( 1 - 2 x ) ^{2}}

Solve

\frac{4 e ^{2} l n}{( 1 - 2 x ) ^{2}} = \frac{e ^{2} l n ( 1 + 2 x ) + 3 ( 1 - 2 x )}{( 1 - 2 x ) ( x - 1 )} : x = \frac{- e ^{2} l n - 3 + 6 e l n ( e ^{2} l n - 1 )}{2 ( e ^{2} l n - 3 )}, x = - \frac{e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3}{2 ( e ^{2} l n - 3 )}; n \neq = \frac{3}{e ^{2} l}

Find the tangent line at

x = \frac{- e ^{2} l n - 3 + 6 e l n ( e ^{2} l n - 1 )}{2 ( e ^{2} l n - 3 )} : y = - \frac{2 ( 3 e ^{4} l ^{2} n ^{2} + 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n - 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3} x - 3 + \frac{2 ( 3 e ^{4} l ^{2} n ^{2} + 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n - 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3}

Find the tangent line at

x = - \frac{e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3}{2 ( e ^{2} l n - 3 )} : y = - \frac{2 ( 3 e ^{4} l ^{2} n ^{2} - 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n + 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n - 6 e l n ( e ^{2} l n - 1 ) + 3} x - 3 + \frac{2 ( 3 e ^{4} l ^{2} n ^{2} - 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n + 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n - 6 e l n ( e ^{2} l n - 1 ) + 3}

y = - \frac{2 ( 3 e ^{4} l ^{2} n ^{2} + 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n - 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3} x - 3 + \frac{2 ( 3 e ^{4} l ^{2} n ^{2} + 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n - 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n + 6 e l n ( e ^{2} l n - 1 ) + 3} or y = - \frac{2 ( 3 e ^{4} l ^{2} n ^{2} - 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n + 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n - 6 e l n ( e ^{2} l n - 1 ) + 3} x - 3 + \frac{2 ( 3 e ^{4} l ^{2} n ^{2} - 6 e ^{3} l n l n e ^{2} l n - 1 - 12 e ^{2} l n + 3 6 e l n ( e ^{2} l n - 1 ) + 9 )}{- 3 e ^{2} l n - 6 e l n ( e ^{2} l n - 1 ) + 3}

x \to 2 lim (\frac{2}{( x + 2 ) ^{2}})

\frac{e ^{x}}{x}

\int tan (x) \cdot sec^{2} (x) d x

\frac{s}{( s - 1 ) ( s ^{2} + 2 s + 2 )}

\frac{d}{d x} (cos^{2} (x) - 2 sin (x))

What is the tangent of f(x)=(1+2x)/(1-2x)enel,(1,-3) ?
The tangent of f(x)=(1+2x)/(1-2x)enel,(1,-3) is