How do you find the derivative of #(tan^3)(4/x)-(cos^-1)(Lnx)#?

Question

1408 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-19T15:47:13

#1/x^2(x/sqrt(1-(lnx)^2)-12tan^2(4/x)sec^2(4/x)), x in (1/e, e)#.

To make #ln x# real, #x > 0#.

Further, #lnx# is a cosine value, and so, #ln x in [-1, 1] #, giving

#x in [1/e, e]#.

So, the function is differentiable for #x in (1/e, e)#.

#(tan^3(4/x)-cos^(-1)(lnx))'#

#=3tan^2(4/x)(tan(4/x))'-(-1/sqrt(1-(lnx)^2))(lnx)'#

#=3tan^2(4/x)(sec^2tan(4/x))'-(-1/sqrt(1-(lnx)^2))(1/x)#

#=3tan^2(4/x)sec^2(4/x)(-4/x^2)+1/sqrt(1-(lnx)^2)(1/x)#

#=1/x^2(x/sqrt(1-(lnx)^2)-12tan^2(4/x)sec^2(4/x))#