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

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SUMMARY

The derivative of the function y = √(ln x) is calculated using u-substitution, where u = ln x. The correct differentiation process involves applying the chain rule, resulting in dy/dx = (1/2) * (1/x) * (1/√(ln x)), which simplifies to dy/dx = 1/(2x√(ln x)). This method effectively addresses the challenge of differentiating a composite function involving a logarithm and a square root.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives.
  • Familiarity with the chain rule in differentiation.
  • Knowledge of logarithmic functions and their properties.
  • Experience with u-substitution in calculus.
NEXT STEPS
  • Study the chain rule in detail to enhance differentiation skills.
  • Practice problems involving u-substitution for various functions.
  • Explore the properties of logarithmic functions and their derivatives.
  • Review examples of differentiating composite functions, particularly those involving square roots.
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Students studying calculus, particularly those struggling with derivatives and composite functions, as well as educators looking for effective teaching strategies in differentiation.

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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