Mind blown - probably missing algebra concept - cannot simplify derivative

frank789

Junior Member
Joined
Sep 16, 2017
Messages
58
hi all!

sorry may have been at this too long but ive been blanking on this one for a while now, been coming at it different ways and I cannot work it down!

f(x) = x^2/sqrt(x^2-1)

so what i did for f'(x)

f(x) = x^2 * 1/sqrt(x^2-1)

product and chain rule to:

f'(x) = 2x/sqrt(x^2-1) - x^3/(x^2-1)^(3/2)

I cannot seem for the life of me to algebra this little rascal down a single term and I feel like I should be able to as solving in this state for the critical points is proving....lets use challenging.

Would really appreciate a point in the right direction!
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,139
hi all!

sorry may have been at this too long but ive been blanking on this one for a while now, been coming at it different ways and I cannot work it down!

f(x) = x^2/sqrt(x^2-1)

so what i did for f'(x)

f(x) = x^2 * 1/sqrt(x^2-1)

product and chain rule to:

f'(x) = 2x/sqrt(x^2-1) - x^3/(x^2-1)^(3/2)

I cannot seem for the life of me to algebra this little rascal down a single term and I feel like I should be able to as solving in this state for the critical points is proving....lets use challenging.

Would really appreciate a point in the right direction!
Much easier to apply quotient rule.

\(\displaystyle \frac{d}{dx} \left[\dfrac{g(x)}{h(x}\right] = \ \dfrac{g'(x)*h(x) - g(x)*h'(x)}{[h(x)]^2}\)

g(x) = x^2 → g'(x) = 2x &

h(x) = (x^2 - 1)^(1/2) → h'(x) = x * (x^2 - 1)^(-1/2)

Now continue.....
 

frank789

Junior Member
Joined
Sep 16, 2017
Messages
58
ok I went through product rule

i got

2x(x^2-1)^(1/2) - (x^3)(x^2-1)^(-1/2)
----------------------------------------------
x^2 - 1

then I dived each term by the x^2 - 1 to get

(2x)(x^2 - 1)^(-1/2) - (x^3)(x^2 - 1)^(-3/2)

which if im not mistaken is the same answer I reached before. as im typing this the first few sips of coffee #2 seem to have done the trick

factor out the largest power of (x^2 - 1), which is (-3/2), along with an x to give

x(x^2 - 1)^(-3/2)(2(x^2 - 1) -x^2)

i expanded inside my binomial and collected like terms to x(x^2 - 1)^(-3/2) * (x^2 - 2)

and I guess since I bothered to type this out I may as well verify I would solve x(x^2 - 2) = 0 for the critical points and get the other critical points from (x^2 - 1) ^ (-3/2) = 0 due to domain restrictions.

thanks again for everything and sorry for posting something I ended up figuring out (im pretty sure anyway), but its amazing what someone verifying your right up until that point does for confidence to finish.
 
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