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

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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 $x in (1/e, e)$ .

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

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

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

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

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

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