What is the derivative of ${(ln (x - 4))}^{3}$ $(ln (x-4)) ^ 3$ ?

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Answer 1 · 2016-07-03T08:43:24

The reqd. deri. $= \frac{3 {(ln (x - 4))}^{2}}{x - 4}$ $={3(ln(x-4))^2)/(x-4)$ .

Let $y = {(ln (x - 4))}^{3}$ $y=(ln(x-4))^3$ , & put $t = ln (x - 4)$ $t=ln(x-4)$ , so, $y=t^3............(1)$

So, $y$ is a fun. of $t$ & $t$ of $x$ .

By Chain Rule, then, we have, the reqd. deri. $=dy/dx=dy/dt*dt/dx$
$=d/dt(t^3)*d/dx{ln(x-4)}$ ..........[by $(1)$ ]

= $3t^2*{1/(x-4)}*d/dx(x-4)=(3t^2)/(x-4)*1={3(ln(x-4))^2)/(x-4)$ .

What is the derivative of (ln (x-4)) ^ 3(ln(x−4))3 (ln (x-4)) ^ 3?