Home > Calculus > Differentiating Logarithmic Functions with Base e

How do you find the derivative of #y=lne^x#?

Answer 1

#dy/dx=1#

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#

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Answer 2

Luke Alvarez

#dy/dx=1#

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#

#dy/dx=1#

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Answer 3

Luke Alvarez

To find the derivative of ( y = \ln(e^x) ), we can use the chain rule. The derivative is given by:

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

Therefore, the derivative of ( y = \ln(e^x) ) is ( \frac{dy}{dx} = 1 ).

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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.