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August 6th, 2012, 12:27 PM   #1
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differentiation of the form b^ax

hello i'm new to the concept of differentiation (dy/dx) and i read
y=e^ax should become a*e^ax (for example e^3x --> 3*e^3x)
but then i was stuck when i had something like y=7^3x;
or something like y=b^ax
so i thought lets transfer b^a to like e^A
by doing this i got ln(b^a)*e^ln(b^a)x
z=e^A
ln(z)=A
b^ax --> A*e^Ax
b^ax --> ln(z)*e^ln(z)x
b^ax --> ln(b^a)*e^ln(b^a)x

could someone tell me if this is correct or if there would be an better way to get this ?
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August 6th, 2012, 12:49 PM   #2
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Re: differentiation of the form b^ax

One way to differentiate:

where a and b are constants is to take the natural log of both sides:



Now, implicitly differentiate with respect to x:





Your result is equivalent after simplification.
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August 6th, 2012, 07:54 PM   #3
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Re: differentiation of the form b^ax

Another equivalent method but we don't have to differentiate logarithmic function this time:



It is known that ; use chain rule:





QED and we completely avoided logarithmic differentiation.

Balarka

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August 6th, 2012, 08:01 PM   #4
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Re: differentiation of the form b^ax

Is exponential differentiation somehow to be preferred over logarithmic differentiation?
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August 6th, 2012, 08:16 PM   #5
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Re: differentiation of the form b^ax

Quote:
Originally Posted by MarkFL
Is exponential differentiation somehow to be preferred over logarithmic differentiation?
Nope. Both are equivalent to me. But, I observed that many people in this forum says that they haven't taught logarithmic differentiation yet, but they do know exponential differentiation.

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August 6th, 2012, 08:20 PM   #6
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Re: differentiation of the form b^ax

Ah...I understand!
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