What is the tangent of f(x)=(14)/((1+e^{-x)),\at (0,7)} ?

The tangent of f(x)=(14)/((1+e^{-x)),\at (0,7)} is

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tangent of f(x)=(14)/((1+e^{-x)),\at (0,7)}

tangent of

f (x) = \frac{14}{( 1 + e ^{- x} ) , at ( 0 , 7 )}

y = \frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) x + \frac{14}{( 1 + e ^{- a_{0}} ) , ( 0 , 7 )} - \frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{14}{( 1 + e ^{- a_{0}} ) , ( 0 , 7 )})

Find the slope of

f (x) = \frac{14}{( 1 + e ^{- x} ) , ( 0 , 7 )} : \frac{df ( x )}{d x} = \frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- x} ) ^{2}} (- e^{- 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{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}}))

Find the line with slope m=

\frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}}))

and passing through

(a_{0}, \frac{14}{( 1 + e ^{- a_{0}} ) , ( 0 , 7 )}) : y = \frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) x + \frac{14}{( 1 + e ^{- a_{0}} ) , ( 0 , 7 )} - \frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) a_{0}

y = \frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) x + \frac{14}{( 1 + e ^{- a_{0}} ) , ( 0 , 7 )} - \frac{14}{, ( 0 , 7 )} (- \frac{1}{( 1 + e ^{- a_{0}} ) ^{2}} (- e^{- a_{0}})) a_{0}

\int \frac{1}{( sin ( x ) ) ^{4}} d x

\frac{\partial}{\partial y} (2 x^{3} y + 1)

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

n = 1 \sum \infty (\frac{3}{4})^{n}

\frac{d}{d t} (sin (2 t))

What is the tangent of f(x)=(14)/((1+e^{-x)),\at (0,7)} ?
The tangent of f(x)=(14)/((1+e^{-x)),\at (0,7)} is