Lny & Derivatives: Why is the Lny y'\y?

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Why is the lny y'\y?

could I also say the lny is dy\y?
 
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Neither of these is correct.
The derivative with respect to y of ln y is 1/y. In symbols, d/dy(ln y) = 1/y
The differential of ln y, d(ln y), is dy/y.

Is that clear?
 


I do understand what I should have said is when taking the derivative with respect to x, why is lny = y'/y. Thank you.
 


It is because of the chain rule. Is that what you are asking?
 


In case you're asking why the derivative of the natural logarithm function ln(x) is 1/x, it is because that is how the function was first defined. More precisely, the function ln(x) is defined to be the integral:
\int_1^x \frac{dt}{t}
It was defined as this integral because it was found that this integral has the properties of a logarithm, the base of which was called 'e', or Euler's number. you can read more about 'e', its discovery and its properties in this book.
The fundamental theorem of calculus then gives you the derivative of the function as 1/x, and the chain rule tells you that if you have f(x) = ln(y(x)), the derivative with respect to x is (1/y(x))*y'(x) or written implicitly, y'/y.
If you are taking the differential of the form ln(y), then you may write d(ln(y)) = dy/y.
 
Last edited:


slider142 said:
In case you're asking why the derivative of the natural logarithm function ln(x) is 1/x, it is because that is how the function was first defined. More precisely, the function ln(x) is defined to be the integral:
\int_1^x \frac{dt}{t}
It was defined as this integral because it was found that this integral has the properties of a logarithm, the base of which was called 'e', or Euler's number. you can read more about 'e', its discovery and its properties in this book.
The fundamental theorem of calculus then gives you the derivative of the function as 1/x, and the chain rule tells you that if you have f(x) = ln(y(x)), the derivative with respect to x is (1/y(x))*y'(x) or written implicitly, y'/y.
If you are taking the differential of the form ln(y), then you may write d(ln(y)) = dy/y.

Thank you for the clarification.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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