# What is the derivative of [ln(2-3x)/lnx]^[arccos(2-5x)]?

Aug 22, 2015

${y}^{'} = {\left[\ln \frac{2 - 3}{\ln} x\right]}^{\arccos \left(2 - 5 x\right)} \cdot \left[\frac{5}{\sqrt{1 - {\left(2 - 5 x\right)}^{2}}} \cdot \ln \left(\ln \frac{2 - 3 x}{\ln} x\right) + \frac{1}{\ln} \left(2 - 3 x\right) \cdot \left(- \frac{3}{2 - 3 x} - \ln \frac{2 - 3 x}{x \cdot \ln x}\right) \cdot \arccos \left(2 - 5 x\right)\right]$

#### Explanation:

The most important derivation rule you'll need to use to differentiate this function is the power rule for a variable base and a variable power.

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f {\left(x\right)}^{g} \left(x\right)\right) = f {\left(x\right)}^{g} \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left[\ln \left(f \left(x\right)\right) \cdot g \left(x\right)\right]}$

In your case, you have $f \left(x\right) = \ln \frac{2 - 3 x}{\ln} \left(x\right)$ and $g \left(x\right) = \arccos \left(2 - 5 x\right)$. Now, I will assume that you know the derivative of $\arccos x$

d/dx(arccosx) = -1/(sqrt(1-x^2)

Ok, buckle up because the calculations will be $\textcolor{red}{\text{horrendous}}$.

So, start your calculation by writing the derivative of $y = f {\left(x\right)}^{g} \left(x\right)$ using the power rule. To keep the calculations as compact as possible, I'll use $f \left(x\right) = f$ and $g \left(x\right) = g$ from this point on.

$\frac{d}{\mathrm{dx}} \left(y\right) = {f}^{g} \cdot \frac{d}{\mathrm{dx}} \left(\ln \left(f\right) \cdot g\right) \text{ } \textcolor{p u r p \le}{\left(1\right)}$

Use the product rule to find

$\frac{d}{\mathrm{dx}} \left(\ln \left(f\right) \cdot g\right) = \left[\frac{d}{\mathrm{dx}} \ln \left(f\right)\right] \cdot g + \ln \left(f\right) \cdot \frac{d}{\mathrm{dx}} \left(g\right) \text{ } \textcolor{p u r p \le}{\left(2\right)}$

Now break this calculation into two different ones. The first one will be

$\frac{d}{\mathrm{dx}} \ln \left(f\right) = \frac{d}{\mathrm{dx}} \left[\ln \left(\ln \frac{2 - 3 x}{\ln} x\right)\right]$

Use the quotient rule once and the chain rule twice, once for $\ln \left(u\right)$, with $u = \ln \frac{2 - 3 x}{\ln} x$, and once more for $\ln \left(v\right)$, with v = 2-3x).

$\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{d}{\mathrm{du}} \ln \left(u\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right) \text{ } \textcolor{p u r p \le}{\left(3\right)}$

$\frac{d}{\mathrm{dx}} \left(u\right) = \frac{\left[\frac{d}{\mathrm{dx}} \ln \left(2 - 3 x\right)\right] \cdot \ln x - \ln \left(2 - 3 x\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right)}{\ln x} ^ 2 \text{ } \textcolor{p u r p \le}{\left(4\right)}$

Use the second chain rule substitution to get

$\frac{d}{\mathrm{dx}} \left(\ln v\right) = \frac{d}{\mathrm{dv}} \ln v \cdot \frac{d}{\mathrm{dx}} \left(v\right)$

$\frac{d}{\mathrm{dx}} \left(\ln v\right) = \frac{1}{v} \cdot \frac{d}{\mathrm{dx}} \left(2 - 3 x\right)$

$\frac{d}{\mathrm{dx}} \left(\ln \left(2 - 3 x\right)\right) = \frac{1}{2 - 3 x} \cdot \left(- 3\right)$

Plug this back into $\textcolor{p u r p \le}{\left(4\right)}$ to get

$\frac{d}{\mathrm{dx}} \left(u\right) = \frac{- \frac{3}{2 - 3 x} \cdot \ln x - \ln \left(2 - 3 x\right) \cdot \frac{1}{x}}{\ln} ^ 2 x$

Now take this result back to $\textcolor{p u r p \le}{\left(3\right)}$ to get

$\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot \frac{- \frac{3}{2 - 3 x} \cdot \ln x - \ln \left(2 - 3 x\right) \cdot \frac{1}{x}}{\ln} ^ 2 x$

$\frac{d}{\mathrm{dx}} \left(\ln \left(f\right)\right) = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\ln x}}}}{\ln} \left(2 - 3 x\right) \cdot \frac{- \frac{3}{2 - 3 x} \cdot \ln x - \ln \left(2 - 3 x\right) \cdot \frac{1}{x}}{\ln} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x$

$\frac{d}{\mathrm{dx}} \left(\ln f\right) = \frac{1}{\ln} \left(2 - 3 x\right) \cdot \left(- \frac{3}{2 - 3 x} - \ln \frac{2 - 3 x}{x \cdot \ln x}\right)$

From $\textcolor{p u r p \le}{\left(2\right)}$, focus on finding $\frac{d}{\mathrm{dx}} \left(g\right)$

$\frac{d}{\mathrm{dx}} \left(g\right) = \frac{d}{\mathrm{dx}} \left(\arccos \left(2 - 5 x\right)\right)$

You're going to have to use th chain rule once for $\arccos t$, with t = 2-5x)

$\frac{d}{\mathrm{dx}} \left(\arccos t\right) = \frac{d}{\mathrm{dt}} \arccos t \cdot \frac{d}{\mathrm{dx}} \left(t\right)$

$\frac{d}{\mathrm{dx}} \arccos t = - \frac{1}{\sqrt{1 - {t}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(2 - 5 x\right)$

$\frac{d}{\mathrm{dx}} \left(\arccos \left(2 - 5 x\right)\right) = - \frac{1}{\sqrt{1 - {\left(2 - 5 x\right)}^{2}}} \cdot \left(- 5\right)$

$\frac{d}{\mathrm{dx}} \left(g\right) = \frac{5}{\sqrt{1 - {\left(2 - 5 x\right)}^{2}}}$

Take everyting back to $\textcolor{p u r p \le}{\left(2\right)}$ to get

$\frac{d}{\mathrm{dx}} \left(\ln \left(f\right) \cdot g\right) = \frac{1}{\ln} \left(2 - 3 x\right) \cdot \left(- \frac{3}{2 - 3 x} - \ln \frac{2 - 3 x}{x \cdot \ln x}\right) \cdot \arccos \left(2 - 5 x\right) + \ln \left(\ln \frac{2 - 3 x}{\ln} x\right) \cdot \frac{5}{\sqrt{1 - {\left(2 - 5 x\right)}^{2}}}$

Finally, take this back to $\textcolor{p u r p \le}{\left(1\right)}$ to get

${y}^{'} = \textcolor{g r e e n}{{\left[\ln \frac{2 - 3}{\ln} x\right]}^{\arccos \left(2 - 5 x\right)} \cdot \left[\frac{5}{\sqrt{1 - {\left(2 - 5 x\right)}^{2}}} \cdot \ln \left(\ln \frac{2 - 3 x}{\ln} x\right) + \frac{1}{\ln} \left(2 - 3 x\right) \cdot \left(- \frac{3}{2 - 3 x} - \ln \frac{2 - 3 x}{x \cdot \ln x}\right) \cdot \arccos \left(2 - 5 x\right)\right]}$