Derivative of f(x)=ln(e^6x^2-4x+5)-e^ln(7x-8)+12e
- Thread starter nell
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D
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nell said:
Just differentiate term by term - using chain rule - i.e.
\(\displaystyle \frac{d}{dx}[f[g(x)] \, = \, \frac {df}{dg} \cdot \frac{dg}{dx}\)
also use
\(\displaystyle e^{ln(a)} \, = \, a\)
and
\(\displaystyle ln[e^a] \, = \, a\)
Please show us your work, indicating exactly where you are stuck, so that we know where to begin to help you.
D
Deleted member 4993
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nell said:
I am confused. My professor has the class using the generalized power rule rather than the chain rule. So, I am unfamiliar with working out the problem as you suggested. I tried it again, and I came up with the following:
f'(x)=12x-4+7-8
f'(x)=12x-5
As always, your assistance is appreciated.
Thanks
f'(x)=12x-4+7-8
f'(x)=12x-5
As always, your assistance is appreciated.
Thanks
mmm4444bot
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nell said:
D
Deleted member 4993
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nell said:
Does your problem look like:
\(\displaystyle f(x) \, = \, ln[e^{6x^2 \, - 4x \, +5}] \, - \, e^{ln(7x-8)} \, + \, 12e\)
This should be written as (in ASCII format):
f(x)=ln(e^(6x^2-4x+5))-e^ln(7x-8)+12e
or does it look like:
\(\displaystyle f(x) \, = \, ln[e^{6x^2} \, - 4x \, +5] \, - \, e^{ln(7x-8)} \, + \, 12e\)
{this is the way wrote it}
D
Deleted member 4993
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Then use the hints I gave you to simplify the expression first
\(\displaystyle e^{ln(a)} \, = \, a\)
and
\(\displaystyle ln(e^a) \, = \, a\)
- then take derivative.
\(\displaystyle e^{ln(a)} \, = \, a\)
and
\(\displaystyle ln(e^a) \, = \, a\)
- then take derivative.