\(\displaystyle y = \sin \sqrt \ln x\)
\(\displaystyle y' = \cos \sqrt \ln x \dfrac{1}{2x \sqrt \ln x} dx\)
The \(\displaystyle \cos \sqrt \ln x \) part of the answer makes sense, but not the other part.
Perhaps some reasoning behind it would be:
\(\displaystyle \sqrt\ln x = (\ln x)^{1/2}\) or
Now substituting u for \(\displaystyle \ln x\)
so the derivative of that would be:
\(\displaystyle \dfrac{1}{2}{(u)^{-1/2}} * \dfrac{1}{x} \)
which can be rewritten as:
\(\displaystyle \dfrac{1}{2x \sqrt u}\)
Now putting \(\displaystyle \ln x\) back into u
\(\displaystyle \dfrac{1}{2x \sqrt \ln x}\)
So the key was understanding that the square root of a number is = to a number to the 1/2 power.
Now, if this was
\(\displaystyle y = \sin \ln x\)
Then
\(\displaystyle y' = \cos \ln x \dfrac{1}{x} dx\)
Is that right?
\(\displaystyle y' = \cos \sqrt \ln x \dfrac{1}{2x \sqrt \ln x} dx\)
The \(\displaystyle \cos \sqrt \ln x \) part of the answer makes sense, but not the other part.
Perhaps some reasoning behind it would be:
\(\displaystyle \sqrt\ln x = (\ln x)^{1/2}\) or
Now substituting u for \(\displaystyle \ln x\)
so the derivative of that would be:
\(\displaystyle \dfrac{1}{2}{(u)^{-1/2}} * \dfrac{1}{x} \)
which can be rewritten as:
\(\displaystyle \dfrac{1}{2x \sqrt u}\)
Now putting \(\displaystyle \ln x\) back into u
\(\displaystyle \dfrac{1}{2x \sqrt \ln x}\)
So the key was understanding that the square root of a number is = to a number to the 1/2 power.
Now, if this was
\(\displaystyle y = \sin \ln x\)
Then
\(\displaystyle y' = \cos \ln x \dfrac{1}{x} dx\)
Is that right?
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