Differerntiating Exponential Functions

[evan4888]

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:

$$y= ( \ln x ) ^\cos x$$

 

[dextercioby]

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

Daniel.

 

[Nylex]

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, $$\frac{dy}{dx} = y\ln a = a^x \ln a$$

 

[evan4888]

Is this even close to the right answer? Or do I need to also multiply by the derivative of the exponent?


$$( \ln x )^\cos x$$ $$( \ln ( \ln x) )$$

 

[dextercioby]

I misses a few terms.You have to use the Leibniz rule of differetiantion of products of functions.

Daniel.

 

[James Jackson]

Nylex, question asked is the differential of

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

not

$$y=\ln{\left (x^\cos{x} \right )}$$

 

[dextercioby]

He didn't go for the function in the OP.He exemplified what is meant by logarithmic differentiation.

Daniel.

 

[James Jackson]

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.

 

[lurflurf]

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

 

[leon1127]

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)...

 

[lurflurf]

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.