What is the tangent of 7/(1-t) ?

The tangent of 7/(1-t) is y= 7/((1-a_{0))^2}t+(-14a_{0}+7)/((-a_{0)+1)^2}

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tangent of 7/(1-t)

tangent of

\frac{7}{1 - t}

y = \frac{7}{( 1 - a _{0} ) ^{2}} t + \frac{- 14 a _{0} + 7}{( - a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

t = a_{0}

Find the tangent point:

(a_{0}, \frac{7}{1 - a _{0}})

Find the slope of

y = \frac{7}{1 - t} : \frac{d y}{d t} = \frac{7}{( 1 - t ) ^{2}}

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{7}{( 1 - a _{0} ) ^{2}}

Find the line with slope m=

\frac{7}{( 1 - a _{0} ) ^{2}}

and passing through

(a_{0}, \frac{7}{1 - a _{0}}) : y = \frac{7}{( 1 - a _{0} ) ^{2}} t + \frac{- 14 a _{0} + 7}{( - a _{0} + 1 ) ^{2}}

y = \frac{7}{( 1 - a _{0} ) ^{2}} t + \frac{- 14 a _{0} + 7}{( - a _{0} + 1 ) ^{2}}

\frac{\partial}{\partial x} (t e^{- t})

\frac{\partial}{\partial x} (2 x^{2} - 8 x + y^{2} - 8 y + 7)

\frac{d}{d x} (\frac{ln ( x ^{3} - y + z )}{x y + z})

\int - 2 x + 2 d x

f (x) = 7 x - ln (x)

What is the tangent of 7/(1-t) ?
The tangent of 7/(1-t) is y= 7/((1-a_{0))^2}t+(-14a_{0}+7)/((-a_{0)+1)^2}