Find dy/dx of y= the square root of ln x

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Homework Help Overview

The discussion revolves around finding the derivative dy/dx for the function y = √(ln x), which falls under the subject area of calculus, specifically differentiation techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of u-substitution to simplify the differentiation process. Some express difficulty in manipulating the expression under the square root and question the validity of their approaches. Others suggest differentiating both sides of an equation derived from squaring y.

Discussion Status

The discussion includes various attempts to differentiate the function, with some participants providing guidance on using u-substitution. There is an acknowledgment of different strategies being considered, but no explicit consensus on the best approach has been reached.

Contextual Notes

Some participants mention struggling with calculus concepts and express a need for clarification on the differentiation process. There is also a reference to the rules of logarithmic differentiation that may be under discussion.

Interception
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Homework Statement


Find dy/dx when y=\sqrt{ln x}


Homework Equations


d/dx of ln x is equal to 1/x times d/dx of x.


The Attempt at a Solution


I tried to raise the ln x to the 1/2 power instead of keeping it under a square root sign, but I had no luck. I'm struggling with Calculus. I would very much appreciate some help.
 
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To visualize the problem in an easier way, let u = ln x. Then proceed. First, differentiate the power, then the term/s under the square root.
 
Interception said:
I tried to raise the ln x to the 1/2 power instead of keeping it under a square root sign, but I had no luck.

This strategy for solving the problem won't work because (ln(x))^a ≠ln(x^a) = a*ln(x). As suggested by sharks, u substitution is the best strategy here.
 
Using u-sub:


\frac{dy}{dx}\sqrt{lnx} | u=lnx

=\sqrt{u}'*u'

=\frac{1}{2\sqrt{lnx}}*\frac{1}{x}

=\frac{1}{2x\sqrt{lnx}}

It's been a while since I've done derivatives, but I think this should be the correct way of working it.
 
There's a nice trick at hand.

y^2 (x) = \ln x

Differentiate both sides with respect to x.
 

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