If Y=tan^-1(2^x / 1+2^x+1), then calculate dy / dx at x=0

At x=0,we have,tany=1/3.

(1+2^x+1)tany=2^x

(1+2^x+1)sec²y.y'+(2^x+1)ln2.tany=2^xln2

(3)(10/9)y'+2ln2(1/3)=ln2

(10/3)y'=(1/3)ln2

y'=ln2/10.

let f=2^x/1+2^x+1

dy/dx=df/dx(1/1+f²)

f(1+2^x+1)=2^x

df/dx(1+2^x+1)+f(2^xlog2)=2^xlog2

find the value df/dx=2(log2)/9

f at x=1 is 1/3

dy/dx=(log2)/5

cheeerssssss.

Thank u .AllamRaju

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