Differerntiating Exponential Functions

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    Exponential Functions
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SUMMARY

The discussion focuses on differentiating the function y = (ln x) ^ cos x using logarithmic differentiation techniques. Participants emphasize the importance of applying the chain rule and the Leibniz rule for differentiating products of functions. The correct differentiation formula is highlighted as (u^v)' = v*(u^(v-1))*u' + (u^v)*log(u)*v', which is essential for solving the problem accurately. The conversation also touches on the complexity of bonus questions in calculus exams, suggesting that such topics should not be considered trivial.

PREREQUISITES
  • Understanding of logarithmic differentiation
  • Familiarity with the chain rule in calculus
  • Knowledge of the Leibniz rule for differentiating products
  • Basic concepts of limits and exponential functions
NEXT STEPS
  • Study the application of logarithmic differentiation in various functions
  • Learn about the Leibniz rule for differentiating products of functions
  • Explore advanced differentiation techniques, including implicit differentiation
  • Review the properties and applications of limits in calculus
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Students studying calculus, particularly those preparing for exams involving differentiation techniques, as well as educators looking to enhance their teaching methods in advanced calculus topics.

evan4888
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I really really need help with this one. This was a bonus question for one of my previous exams. I have no idea how to work through it.

Differentiate the function:

y= ( \ln x ) ^\cos x
 
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Do you know about logarithmic differentiation...?If you don't,are you willing to learn? :smile:

Daniel.
 
Logarithmic differentiation:

y = a^x

\ln y = \ln (a^x) = x\ln a

Then by the chain rule: \frac{1}{y} \frac{dy}{dx} = \ln a (only true if a is a constant. If a = a(x), then you need to apply the chain and product rules to the RHS).

So, \frac{dy}{dx} = y\ln a = a^x \ln a
 
Is this even close to the right answer? Or do I need to also multiply by the derivative of the exponent?


( \ln x )^\cos x ( \ln ( \ln x) )
 
I misses a few terms.You have to use the Leibniz rule of differetiantion of products of functions.

Daniel.
 
Nylex, question asked is the differential of

y=(\ln{x})^\cos{x}

not

y=\ln{\left (x^\cos{x} \right )}
 
He didn't go for the function in the OP.He exemplified what is meant by logarithmic differentiation.

Daniel.
 
True, but it was implied that you can use the \ln{\left (x^a\right )}=a\ln{x} rule directly for the problem at hand.
 
In all The usual formulas elementary calculus students are expected to learn why is
(u^v)'=v*(u^(v-1))*v'+(u^v)*log(u)*v'
Rarely included. It is quite nice and it is easy to see using the chain rule.
Or Individual function differential operators.
D(u^v)=(Du+Dv)(u^v)=(Du)(u^v)+(Dv)(u^v)
where (Du)f(u,v)=Dx f(u(x),v(y))|y=x
(Dv)f(u,v)=Dx f(u(y),v(x))|y=x
Dx y=0
 
  • #10
evan4888 said:
I really really need help with this one. This was a bonus question for one of my previous exams. I have no idea how to work through it.

Differentiate the function:

y= ( \ln x ) ^\cos x

If this is the bonus on the exams... your teacher is really nice..

my bonus question on the test was
Limit(x^x, x, 0)...
 
  • #11
leon1127 said:
If this is the bonus on the exams... your teacher is really nice..

my bonus question on the test was
Limit(x^x, x, 0)...
These should not be bonus questions.
x^x:=exp(x log(x))
x^x~1+x log(x)~1+|x|
Also le Hopitals rule works nicely.
What I want to know is did you mean the directed limit? You need complex numbers to consider x^x for x<0.
 

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