# How do you differentiate  f(x)=sqrt((ln(-x+3))^3 using the chain rule.?

May 8, 2017

(df)/(dx)=-(3(ln(3-x))^2)/(2(3-x)sqrt((ln(3-x))^3)

#### Explanation:

Chain Rule - In order to differentiate a function of a function, say $y , = f \left(g \left(x\right)\right)$, where we have to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we need to do (a) substitute $u = g \left(x\right)$, which gives us $y = f \left(u\right)$. Then we need to use a formula called Chain Rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$. In fact if we have something like $y = f \left(g \left(h \left(k \left(x\right)\right)\right)\right)$, we can have $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{df}} \times \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dk}}{\mathrm{dx}}$

Here we have $f \left(x\right) = \sqrt{{\left(\ln \left(- x + 3\right)\right)}^{3}}$ or say

$f \left(x\right) = \sqrt{g \left(x\right)}$, where $g \left(x\right) = {\left(h \left(x\right)\right)}^{3}$, $h \left(x\right) = \ln \left(k \left(x\right)\right)$ and $k \left(x\right) = - x + 3$

and (df)/(dg)=1/(2sqrt(g(x)), $\frac{\mathrm{dg}}{\mathrm{dh}} = 3 \left(h {\left(x\right)}^{2}\right)$, $\frac{\mathrm{dh}}{\mathrm{dk}} = \frac{1}{k \left(x\right)}$ and $\frac{\mathrm{dk}}{\mathrm{dx}} = - 1$

hence (df)/(dx)=1/(2sqrt(g(x))xx3(h(x)^2)xx1/(k(x))xx(-1)

= $\frac{1}{2 \sqrt{{\left(\ln \left(- x + 3\right)\right)}^{3}}} \times 3 {\left(\ln \left(- x + 3\right)\right)}^{2} \times \frac{1}{\left(- x + 3\right)} \times \left(- 1\right)$

= -(3(ln(-x+3))^2)/(2(-x+3)sqrt((ln(-x+3))^3)

= -(3(ln(3-x))^2)/(2(3-x)sqrt((ln(3-x))^3)