What is the tangent of y=7ln((e^x+e^{-x})/2),(0,0) ?

The tangent of y=7ln((e^x+e^{-x})/2),(0,0) is y=0

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tangent of y=7ln((e^x+e^{-x})/2),(0,0)

tangent of

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

y = 0

Solution steps

Find the slope of

y = 7 ln (\frac{e ^{x} + e ^{- x}}{2}) : \frac{d y}{d x} = \frac{7 ( e ^{2 x} - 1 )}{e ^{2 x} + 1}

Compute the slope of

y = 7 ln (\frac{e ^{x} + e ^{- x}}{2})

(0, 0) : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

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

\int 15 x + \frac{3}{x} d x

y = x^{2} + 2 x + 1

x \frac{d y}{d x} = y^{2} + 1

\int (\frac{1}{4 + x ^{2}}) d x

What is the tangent of y=7ln((e^x+e^{-x})/2),(0,0) ?
The tangent of y=7ln((e^x+e^{-x})/2),(0,0) is y=0