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

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Answer 1 · 2016-12-30T06:52:06

See explanation.

For this function to be real,

$x > 0, 2 - 3 x > 0 and - 1 \leq 2 - 5 x \leq 1$ $x>0, 2-3x>0 and -1<=2-5x<=1$

compounding,

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

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

the ratio is positive.

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

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

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

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

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

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

$(arc cos (2-5x)$

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

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

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