What is the derivative of #f(x)=sin(1/lnx)#?

Answer 1

#f'(x) = -(cos((1)/(ln x)))/(xln^2x)#

We need to use the chain rule.

First, we can rename the function to make the calculations easier. So, we have a few functions called:

#f(u) = sin(u)#
where #u(t) = (1)/(t)#
and #t(x) = ln x#
So, the derivative of #f(x)# is (using the chain rule):
#f'(x) = (df)/(du)(du)/(dt)(dt)/(dx) #

We calculate these derivatives:

#(df)/(du) = cos(u)#
#(du)/(dt) = -(1)/(t^2)#
#(dt)/(dx) = (1)/(x)#

The final expression is the multiplication of them and the substitution of their previous definitions:

#f'(x) = cos(u)(-1)/(t^2)(1)/(x) = -cos((1)/(lnx))(1)/(xln^2x)#
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Answer 2

To find the derivative of ( f(x) = \sin(\frac{1}{\ln x}) ), we apply the chain rule:

[ f'(x) = \cos(\frac{1}{\ln x}) \cdot \frac{d}{dx}(\frac{1}{\ln x}) ]

Using the chain rule, the derivative of ( \frac{1}{\ln x} ) with respect to ( x ) is:

[ \frac{d}{dx}(\frac{1}{\ln x}) = -\frac{1}{(\ln x)^2} \cdot \frac{d}{dx}(\ln x) ]

[ = -\frac{1}{(\ln x)^2} \cdot \frac{1}{x} ]

Putting it all together:

[ f'(x) = \cos(\frac{1}{\ln x}) \cdot (-\frac{1}{(\ln x)^2} \cdot \frac{1}{x}) ]

[ = -\frac{\cos(\frac{1}{\ln x})}{x(\ln x)^2} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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