Question

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

Answer 1

See explanation.

Explanation:

For this function to be real,

`x>0, 2-3x>0 and -1<=2-5x<=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 =(ln(2-3x)/ln x)^(arc cos (2-5x))`.

`ln y =arc cos(2-5x)(ln ln(2-3x)-ln ln x)`. Differentiating,

`1/yy'=(arc cos (2-5x) )`

`(((-3))/(ln(2-3x)(2-3x))-1/((ln x)(x)))`

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

`y'=-(ln(2-3x)/ln x)^(arc cos (2-5x))`

`(arc cos (2-5x)`

`(3/(ln(2-3x)(2-3x))+1/((ln x)(x)))`

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