What is the derivative of tan((1-e^x/(1+e^x))) ?

The derivative of tan((1-e^x/(1+e^x))) is -(2e^xsec^2(\frac{1-e^x)/(1+e^x))}{(1+e^x)^2}

What is the first derivative of tan((1-e^x/(1+e^x))) ?

The first derivative of tan((1-e^x/(1+e^x))) is -(2e^xsec^2(\frac{1-e^x)/(1+e^x))}{(1+e^x)^2}

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\frac{d}{d x} (tan (\frac{1 - e ^{x}}{1 + e ^{x}}))

- \frac{2 e ^{x} sec ^{2} ( \frac{1 - e ^{x}}{1 + e ^{x}} )}{( 1 + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (tan (\frac{1 - e ^{x}}{1 + e ^{x}}))

Apply the chain rule:

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

= sec^{2} (\frac{1 - e ^{x}}{1 + e ^{x}}) \frac{d}{d x} (\frac{1 - e ^{x}}{1 + e ^{x}})

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

= sec^{2} (\frac{1 - e ^{x}}{1 + e ^{x}}) (- \frac{2 e ^{x}}{( 1 + e ^{x} ) ^{2}})

Simplify

sec^{2} (\frac{1 - e ^{x}}{1 + e ^{x}}) (- \frac{2 e ^{x}}{( 1 + e ^{x} ) ^{2}}) : - \frac{2 e ^{x} sec ^{2} ( \frac{1 - e ^{x}}{1 + e ^{x}} )}{( 1 + e ^{x} ) ^{2}}

= - \frac{2 e ^{x} sec ^{2} ( \frac{1 - e ^{x}}{1 + e ^{x}} )}{( 1 + e ^{x} ) ^{2}}

x \to 0 lim (\frac{1}{cos ( 5 x )})

f (x) = e^{x + 1}

x \to 6 lim (\frac{( \frac{1}{x} - \frac{1}{6} )}{x - 6})

\frac{d}{d x} (arccos (3 x) - 2 sinh (\frac{x}{5}))

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

What is the derivative of tan((1-e^x/(1+e^x))) ?
The derivative of tan((1-e^x/(1+e^x))) is -(2e^xsec^2(\frac{1-e^x)/(1+e^x))}{(1+e^x)^2}
What is the first derivative of tan((1-e^x/(1+e^x))) ?
The first derivative of tan((1-e^x/(1+e^x))) is -(2e^xsec^2(\frac{1-e^x)/(1+e^x))}{(1+e^x)^2}