This is a problem of differentiation with infinite series...
well am not getting simply 2 as answer. My answer is 2y^2.
part of my solution is,
let,
y = ( tan x )^y
logy = y log(tan x)
(1/y)dy/dx = y(1/tan x)sec^2x + log(tan x)dy/dx
(1/y)dy/dx = y/(sin x)(cos x)
dy/dx = 2y^2
it should be , dy/dx=2
plzzzz... help
well am not getting simply 2 as answer. My answer is 2y^2.
part of my solution is,
let,
y = ( tan x )^y
logy = y log(tan x)
(1/y)dy/dx = y(1/tan x)sec^2x + log(tan x)dy/dx
(1/y)dy/dx = y/(sin x)(cos x)
dy/dx = 2y^2
it should be , dy/dx=2
plzzzz... help
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You are almost there:
Now y = tan x so you have dy/dx = 2(tan x)^2
At x = pi/4, tan x = 1.
Substitute to get 2(1)^2 which is equal to 2.
Now y = tan x so you have dy/dx = 2(tan x)^2
At x = pi/4, tan x = 1.
Substitute to get 2(1)^2 which is equal to 2.