How do you find the first and second derivative of $ln ({ln x}^{2})$ $ln(lnx^2)$ ?

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Answer 1 · 2016-08-18T13:32:06

$(1) : \frac{d y}{d x} = \frac{1}{x ln x}$ $(1) : dy/dx=1/(xlnx)$ .

$(2) : \frac{d^{2} y}{{d x}^{2}} = - \frac{1 + ln x}{x^{2} {(ln x)}^{2}}$ $(2) : (d^2y)/dx^2=-(1+lnx)/(x^2(lnx)^2)$ .

Let $y = ln ({ln x}^{2})$ $y=ln(lnx^2)$

Using the well-known Rules of Log. Fun., we have,

$y = ln (2 ln x) = ln 2 + ln (ln x)$ $y=ln(2lnx)=ln2+ln(lnx)$

By the Chain Rule, then,

$\frac{d y}{d x} = \frac{d}{d x} (ln 2) + \frac{d}{d x} {ln (ln x)}$ $dy/dx=d/dx(ln2)+d/dx{ln(lnx)}$

$= 0 + \frac{1}{ln x} \cdot \frac{d}{d x} (ln x)$ $=0+1/lnx*d/dx(lnx)$

$:. dy/dx=1/lnx*1/x=1/(xlnx)$ .

Before proceeding further for the $2^(nd)$ Derivative $(d^2y)/dx^2$ , let us recall that $d/dt(1/t)=-1/t^2$ . Hence,

$(d^2y)/dx^2=d/dx(dy/dx)=d/dx{1/(xlnx)}$

$=-1/((xlnx)^2)*d/dx{xlnx}$

$=-1/(x^2(lnx)^2){x*d/dx(lnx)+(lnx)*d/dx(x)}$

$=-1/(x^2(lnx)^2){x*1/x+(lnx)(1)}$

$:. (d^2y)/dx^2=-(1+lnx)/(x^2(lnx)^2)$ .

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How do you find the first and second derivative of ln(lnx^2)ln(lnx2)ln(lnx^2)?