Differentiate with respect to x, `y = e^tanx `

Explanation

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a. We know that
`2tan^(-1)x={{:(sin^(-1)((2x)/(1+x^(2)))",",if -1lexle1),(pi-sin^(-1)((2x)/(1+x^(2)))",",if xgt 1),(-pi-sin^(-1)((2x)/(1+x^(2)))",", if x lt -1):}`
`"or "(dy)/(dx)=-(2)/(1+x^(2))if x lt -1 or x gt 1`
b. `cos^(-1)((1)/(sqrt(1+x^(2))))={{:(tan^(-1)x",", xge0),(-tan^(-1)x",",x lt0):}`
`"or "(dy)/(dx)=-(1)/(1+x^(2))if x lt 0`
c. `y=|e|^(|x|)-e|={{:(|e^(x)-e|",", xge0),(|e^(-x)-e|",", xlt0):}={{:(e^(x)-e",",x ge 1),(e-e^(x)",", 0lt x lt 1),(e-e^(-x)",", -1 le x lt 0),(e^(-x)-e",", x lt-1):}`
`"or "(dy)/(dx)gt0if x gt or -1 lt x lt 0.`
d. `u=log |2x|, y=|tan^(-1)x|`
`"or "(du)/(dx)=(1)/(x)and (dv)/(dx)={{:((1)/(1+x^(2))",",x gt0),(-(1)/(1+x^(2))",", xlt 0):}`
`therefore" " (dv)/(dv)={{:((1)/(1+x^(2))",",x gt0),(-(1)/(1+x^(2))",", xlt 0):}`
Now, we know that
`(1+x^(2))/(x)=x+(1)/(x)gt2 if x gt1 and lt -2 if x lt -1`
`therefore" "(du)/(dv)gt2 if xlt-1 or xgt1`