Differerntiating Exponential Functions

  • Thread starter 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:

[tex]y= ( \ln x ) ^\cos x[/tex]
 

dextercioby

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Do you know about logarithmic differentiation...?If you don't,are you willing to learn? :smile:

Daniel.
 
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Logarithmic differentiation:

[tex]y = a^x[/tex]

[tex]\ln y = \ln (a^x) = x\ln a[/tex]

Then by the chain rule: [tex]\frac{1}{y} \frac{dy}{dx} = \ln a[/tex] (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, [tex]\frac{dy}{dx} = y\ln a = a^x \ln a[/tex]
 
Is this even close to the right answer? Or do I need to also multiply by the derivative of the exponent?


[tex] ( \ln x )^\cos x [/tex] [tex] ( \ln ( \ln x) ) [/tex]
 

dextercioby

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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

[tex]y=(\ln{x})^\cos{x}[/tex]

not

[tex]y=\ln{\left (x^\cos{x} \right )}[/tex]
 

dextercioby

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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 [itex]\ln{\left (x^a\right )}=a\ln{x}[/itex] rule directly for the problem at hand.
 

lurflurf

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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
 
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:

[tex]y= ( \ln x ) ^\cos x[/tex]
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)...
 

lurflurf

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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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