Differentiate the following function with respect to `x:` `(logx)^(cosx)`

Explanation

Let y=`(logx) ^cosx`

Taking logarithm on both the sides, we obtain

logy=cosx.log(logx) ..... `[∵loga^x =xloga]`

Differentiating both sides with respect to x, we obtain

`1/y.(dx)/(dy)=d/(dx)(cosx)×log(logx)+cosx× d/(dx) [log(logx)]`

⇒`1/y.(dx)/(dy)=−sinxlog(logx)+cosx× 1/(logx).​d/(dx)​ (logx)`

⇒ `(dx)/(dy)​=y[−sinxlog(logx)+(cosx)/(logx)​ × 1/x ]`

∴`(dy)/(dx) =(logx)^cosx [ (cos x)/(xlogx) −sinxlog(logx)]`