How do you find the derivative of (tan^3)(4/x)-(cos^-1)(Lnx)(tan3)(4x)−(cos−1)(lnx)?

1 Answer
A. S. Adikesavan · Monzur R.
Feb 19, 2017

1/x^2(x/sqrt(1-(lnx)^2)-12tan^2(4/x)sec^2(4/x)), x in (1/e, e)1x2⎛⎜ ⎜⎝x√1−(lnx)2−12tan2(4x)sec2(4x)⎞⎟ ⎟⎠,x∈(1e,e).

Explanation:

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

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

x in [1/e, e]x∈[1e,e].

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

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