What is the derivative of tan^4((1-x/(1+x))) ?

The derivative of tan^4((1-x/(1+x))) is -(8tan^3(\frac{1-x)/(1+x))sec^2((1-x)/(1+x))}{(1+x)^2}

What is the first derivative of tan^4((1-x/(1+x))) ?

The first derivative of tan^4((1-x/(1+x))) is -(8tan^3(\frac{1-x)/(1+x))sec^2((1-x)/(1+x))}{(1+x)^2}

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\frac{d}{d x} (tan^{4} (\frac{1 - x}{1 + x}))

- \frac{8 tan ^{3} ( \frac{1 - x}{1 + x} ) sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}

Solution steps

\frac{d}{d x} (tan^{4} (\frac{1 - x}{1 + x}))

Apply the chain rule:

4 (tan (\frac{1 - x}{1 + x}))^{3} \frac{d}{d x} (tan (\frac{1 - x}{1 + x}))

= 4 (tan (\frac{1 - x}{1 + x}))^{3} \frac{d}{d x} (tan (\frac{1 - x}{1 + x}))

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

= 4 (tan (\frac{1 - x}{1 + x}))^{3} (- \frac{2 sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}})

Simplify

4 tan^{3} (\frac{1 - x}{1 + x}) (- \frac{2 sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}) : - \frac{8 tan ^{3} ( \frac{1 - x}{1 + x} ) sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}

= - \frac{8 tan ^{3} ( \frac{1 - x}{1 + x} ) sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}

\frac{d y}{d t} = - 0.4 y + 6.2 e^{- 0.73 t}, y (0) = 0

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{( x ^{2} - 1 )}, y (2) = 2

x \to \infty lim (5 x)

\int_{0}^{6} 7000 t e^{- 0.1 t} d t

\int_{1}^{10} \frac{5}{x ^{3}} d x

What is the derivative of tan^4((1-x/(1+x))) ?
The derivative of tan^4((1-x/(1+x))) is -(8tan^3(\frac{1-x)/(1+x))sec^2((1-x)/(1+x))}{(1+x)^2}
What is the first derivative of tan^4((1-x/(1+x))) ?
The first derivative of tan^4((1-x/(1+x))) is -(8tan^3(\frac{1-x)/(1+x))sec^2((1-x)/(1+x))}{(1+x)^2}