# How do you find the derivative of the function: [ln(2-3x)/lnx]^[arccos(2-5x)]?

Dec 30, 2016

See explanation.

#### Explanation:

For this function to be real,

$x > 0 , 2 - 3 x > 0 \mathmr{and} - 1 \le 2 - 5 x \le 1$

compounding,

all inclusive 1/5<=x<=3/5.

And, for this range, the two lagarithms are negative, and therefore,

the ratio is positive.

Let $y = {\left(\ln \frac{2 - 3 x}{\ln} x\right)}^{a r c \cos \left(2 - 5 x\right)}$.

$\ln y = a r c \cos \left(2 - 5 x\right) \left(\ln \ln \left(2 - 3 x\right) - \ln \ln x\right)$. Differentiating,

$\frac{1}{y} y ' = \left(a r c \cos \left(2 - 5 x\right)\right)$

$\left(\frac{\left(- 3\right)}{\ln \left(2 - 3 x\right) \left(2 - 3 x\right)} - \frac{1}{\left(\ln x\right) \left(x\right)}\right)$

-(-((-5))/sqrt(1-(2-5x)^2)))(ln ln (2-3x)-ln ln x)

$y ' = - {\left(\ln \frac{2 - 3 x}{\ln} x\right)}^{a r c \cos \left(2 - 5 x\right)}$

(arc cos (2-5x)

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

+(5)/sqrt(1-(2-5x)^2)(ln ln (2-3x)-ln ln x)), 1/5<=x<=3/5#