derivative of f(x)=ln((x^(8)-8)^(5)) ??
\[\large f(x)=\ln\left[(x^8-8)^5\right]\]Can you see the formatted text above? So we need to take the derivative of this, yes? :)
Yes! thats it!
Do you recall the derivative of \(\large \ln x\) ?
1/x?
Oh I guess we could apply a rule of logarithms to make this a tad easier.\[\large \log (a^b)=b \cdot \log a\] So before we do anything, let's rewrite our equation like this,\[\large f(x)=5\cdot\ln\left[(x^8-8)\right]\]Using the above mentioned rule to bring the 5 out front.
Yes, very good, 1/x. We'll be using that derivative rule for our first step.
okay sounds good
\[\large f(x)=5\cdot\ln\left(x^8-8\right)\] \[\large f'(x)=5\cdot \frac{1}{x^8-8}\color{royalblue}{\left(x^8-8\right)'}\] Ignore the blue term for now, and ignore the 5. The 5 will just follow us along for the ride. Understand the first step? Taking the derivative of the log, places EVERYTHING inside the log, into the denominator.
okay that makes sense!
The blue term shows up due to the chain rule. The chain rule tells us we have to multiply by the derivative of the inner function. The little prime on the outside is to show that we need to take it's derivative still. What do you get taking the derivative of the blue part?
\[8x ^{7}\]
?
\[\large f'(x)=5\cdot \frac{1}{x^8-8}\color{royalblue}{\left(8x^7\right)}\]Yup sounds good :) To finish up, you could put all that stuff on top of the fraction so it looks a little nicer.\[\large f'(x)=\cdot \frac{5\cdot 8x^7}{x^8-8}=\frac{40x^7}{x^8-8}\]
can anything multiplied by a fraction be moved to the numerator like that?
Yes, here are a couple things to remember.\[\large 5 = \frac{5}{1}\] So when we're multiplying this,\[\large 5\cdot \frac{1}{x^8-8}\] We're really just multiplying fractions,\[\large \frac{5}{1}\cdot \frac{1}{x^8-8}\]
And then from there, you just need to remember how to multiply fractions (top by top, bottom by bottom).
great! so is 40x^7 / 8x^8 the final answer?
yup! good job.
I meant: x^8-8
awesome thanks!!1
hehe that' what i thought u meant XD
thank you!!
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