Calculus I Derivatives of exponential and logarithmic functions

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Evaluate ln e^(2ln x)

Use logarithm rules to move out of the exponent.
Simplify by moving inside the logarithm.
The natural logarithm of is .
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How do you find the derivative of #y=lne^x#?

2 Answers

mason m

Dec 5, 2016

Answer:

#dy/dx=1#

Explanation:

We should know for this approach that #d/dxln(x)=1/x#.

Applying the chain rule to this derivative tells us that if we were to have a function #u# instead of just the variable #x# within the logarithm, we see that #d/dxln(u)=1/u*(du)/dx#.

So we see that #d/dxln(e^x)=1/e^x*d/dx(e^x)#

Since #d/dx(e^x)=e^x#:

#d/dxln(e^x)=1/e^x*e^x=1#

mason m

Dec 5, 2016

Answer:

#dy/dx=1#

Explanation:

The logarithm function and exponential functions are inverse functions–they undo one another! This means that #log_a(a^x)=x# and #a^(log_a(x))=x#.

Recall that the function #ln(x)# is the logarithm with a base of #e#, that is, #ln(x)=log_e(x)#. Thus:

#y=ln(e^x)=log_e(e^x)=x#

So

#dy/dx=1#

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