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How do I find the derivative of #15/2+ln(5x) #?

Answer 1

#1/x#

First, notice that differentiating the #15/2# will just give #0#, so this question is identical to just finding the derivative of #ln(5x)#.

In order to differentiate functions with the natural logarithm, it's necessary to know that #d/dx(ln(x))=1/x#.

Then, through the chain rule, this can be generalized to say that #d/dx(ln(u))=1/u*u'#.

Thus,

#d/dx(ln(5x))=1/(5x)*d/dx(5x)#

Since #d/dx(5x)=5#, the derivative of the original function is

#1/(5x)*5=1/x#

Notice that this derivative is the exact same as the derivative of just #ln(x)#. This can be explained using logarithm rules:

#ln(5x)=ln(5)+ln(x)#

So, #d/dx(ln(5)+ln(x))=0+1/x=1/x#

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Answer 2

Axel Alvarez

To find the derivative of ( \frac{15}{2} + \ln(5x) ), use the following steps:

The derivative of a constant is 0.
The derivative of ( \ln(u) ) with respect to ( x ) is ( \frac{1}{u} \cdot \frac{du}{dx} ).

Applying these rules:

( \frac{d}{dx}(\frac{15}{2} + \ln(5x)) = \frac{d}{dx}(\frac{15}{2}) + \frac{d}{dx}(\ln(5x)) )

( = 0 + \frac{1}{5x} \cdot 5 )

( = \frac{1}{x} )

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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.