SOLUTION: Hello everyone, finding the following question puzzling, i have not covered this before, and am trying to get a heads up for when i start my course
Use logarithmic differentiati
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Question 375065: Hello everyone, finding the following question puzzling, i have not covered this before, and am trying to get a heads up for when i start my course
Use logarithmic differentiation to differentiate the following:
(i) y = 2^x (ii) y = x^x
Cheers
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
i)
Start with the given equation.
Take the natural log of both sides.
Pull down the exponent.
)
... Derive both sides
)
... Solve for y'
)
... Now plug in
Note: the derivative of is
-----------------------------------------------
ii)
Start with the given equation.
Take the natural log of both sides.
Pull down the exponent.
+1)
... Derive both sides
+1))
... Solve for y'
+1))
... Now plug in
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You get one question per post. Since the first one is trivial, I'll do the second.
Take the natural log of both sides:
\ =\ \ln\left(x^x\right))
\ =\ x\ln(x))
Differentiate both sides with respect to 
using the Chain Rule on the left and the Product Rule on the right:
Let )
.
Then 
Hence:
\ =\ x\,\cdot\,\left(\frac{1}{x}\right)\ +\ \ln(x)\ =\ 1\ +\ \ln(x))
Multiply both sides by 
\right))
John
My calculator said it, I believe it, that settles it
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