f(x)=x^9(x-7)^3/(x^2+6)^2 f(x)'=
logarithmic differentiation

Question

f(x)=x^9(x-7)^3/(x^2+6)^2 f(x)'=
logarithmic differentiation

anonymous · Answer

You have y=x^9(x-7)^3/(x^2+6)^2
Now you know ln(y)=ln(x^9(x-7)^3/(x^2+6)^2)
Split up the lograrithm on the right and the proceed to differentiate both sides

anonymous · Answer

ln(f(x))=ln(x^9(x-7)^3)-2ln(x^2+6)
f'(x)/f(x)=9x^8(x-7)^3+3x^9(x-7)^2/x^9(x-7)^3-8x//x^2+6
Now you multiply times f(x) to each side and you already know what f(x) is

anonymous · Answer

$$In(y)=\frac{ In(x)^9(x-7)^3 }{ (x^2+6)^ 2}$$

anonymous · Answer

whats next?

anonymous · Answer

f'(x)=(9x^8(x-7)^3+3x^9(x-7)^2/x^9(x-7)^3-8x//x^2+6)x^9(x-7)^3/(x^2+6)^2

anonymous · Answer

could you put the equation by using the below bottom?

anonymous · Answer

$$\ln(f(x))=\ln(x^9(x-7)^3)-2\ln(x^2+6)$$
$$f'(x)/f(x)=[(9x^8(x-7)^3+3x^9(x-7)^2)/(x^9(x-7)^3]-[4x/x^2+6)]$$
$$f'(x)={[[(9x^8(x-7)^3+3x^9(x-7)^2/x^9(x-7)^3]-[4x/x^2+6)]]} x^9(x-7)^3/(x^2+6)^2$$

anonymous · Answer

Thank you, its clear! but the after x^9(x... cut off!

anonymous · Answer

You're welcome!

anonymous · Answer

after "]]x^9"?

anonymous · Answer

yes :/

anonymous · Answer

It is the function f(x)
$$x^9(x-7)^3/(x^2+6)^2$$
So it is being multiplied to everything.

anonymous · Answer

is there any other easier way to do this?

anonymous · Answer

This is one of the easiest ways because you get to avoid quotient rule and it mainly involves the product rule.

anonymous · Answer

oh I see. thank you  what is the final answer?

anonymous · Answer

all of it but if you notice the part where there is an$$(x-7)^2/(x-7)^3$$ that could be simplified to $$1/(x-7)$$
If you want to simplify further you'll probably have to multiply it out and see if anything gets factored out or cancels.

anonymous · Answer

the answer is 1/(x-7)?

anonymous · Answer

no. That was just a part that could be simplified. The answer is the long part where the x^9 was intially cut off:O

anonymous · Answer

Oh no haha 
x^9(x-7)^3/(x^2+6)^2

anonymous · Answer

?

anonymous · Answer

and everything before that:O It can be a pain to write out lol

anonymous · Answer

wow, thats a really long answer!!

anonymous · Answer

would it be possible to shorten?

anonymous · Answer

It is!!! Well if you try to see if anything cancel or factors out in order to cancel

anonymous · Answer

mmm... x^9 is canceling out?

anonymous · Answer

I don't think so well maybe after a lot of expanding and factoring it will

anonymous · Answer

Ohk.

anonymous · Answer

can you help me with other question?

anonymous · Answer

Post a new question.

anonymous · Answer

Ok!