Question

slove it with clear explanation​

Answer 1

Questions

y = (Tanx)^logx + cos^2(π/4) find dy/dx

Solutions

let

U = (Tanx)^logx

using log

logu = logx .log{tan(X)}

differentiating with respect to X

du/udx = [log{Tan(x)}/X + logx(sec^2x)/tanx]u

Now

y = u + cos^2(π/4)

differentiating with respect to X we get

dy/dx = du/dx + 0

On putting the value of du/dx and u we get

dy/dx =[ log{tan(X)}/X + logx sec^2x/tanx]*{tan(X)}^logx