How do you differentiate $f(x) = ln(sqrt(arcsin(e^(2-x)) )$ using the chain rule?

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Answer 1 · 2016-05-07T14:09:49

$-1/2e^(2-x)/(arcsin(e^(2-x))sqrt(1-e^(4-2x)))$

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Draw a triangle to express your inverse trigonometric term.
We find that $e^(2-x)=siny$ or $arc(e^(2-x))=y$ .

We want to find $(df)/(dx)$ which also equals to $(df)/(dy)(dy)/(dx)$ .

$(df)/(dy)=d/(dy)ln(sqrty)=1/(2y)$

Differentiate explicitly:
$e^(2-x)=siny$
$-e^(2-x)dx=(cosy)dy$ or $(dy)/(dx)=-e^(2-x)/cosy$

Therefore, your answer is;
$(df)/(dx)=(df)/(dy)(dy)/(dx)$ .
$=1/(2y)xx-e^(2-x)/cosy$
$=-1/2e^(2-x)/(arcsin(e^(2-x))sqrt(1-e^(4-2x)))$

Note:
$y=arc(e^(2-x))$
$cosy= sqrt(1-e^(4-2x))$ (refer to the triangle)

Answer 2 · 2016-05-07T14:51:09

Just adding to Alexander's answer, $x>=2$ .. .

$e^(2-x)>=0$ . Yet, as it is sine of an angle, it has to be $<=1$ . So, the exponent $2-x<=0$ . Thus, $x>=2$ .

Also, for differentiation, arc sine should be taken as a single-valued function..

How do you differentiate f(x) = ln(sqrt(arcsin(e^(2-x)) ) f(x) = ln(sqrt(arcsin(e^(2-x)) ) using the chain rule?