C3 help please? differentiation

Hiya guys,
I'm having trouble with a problem form my c3 text book. The question is:
Show that the curve y=(x^3)/(1+x^4)^0.5 has a positive gradient for all values of x except at x=0
I have differentiated the equation to find dy/dx however i cannot see how this will always be positive.
Thanks

Reply 1

L'Evil Fish

What's your dy/dx equation?

Reply 2

Indeterminate

Original post by noodlemon

You should get

\dfrac{dy}{dx} = \dfrac{x^2(x^4+3)}{(1+x^4)^{1.5}}

Now, notice that

x^2 > 0

for all non-zero x. So what can you say about

x^4

(edited 11 years ago)

Reply 3

noodlemon

originally, (3x^2(1+x^4)^0.5)-(2x^6(x^4+1))^0.5)/(x^4+1), which i simplified in the end to, x^2(3+x^4)/(x^4+1)^3/2 ?? i used the quotient rule.

Reply 4

noodlemon

Original post by Indeterminate

You should get

\dfrac{dy}{dx} = \dfrac{x^2(x^4+3)}{(1+x^4)^{1.5}}

Now, notice that

x^2 > 0

for all non-zero x. So what can you say about

x^4

yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?

Reply 5

ThatRandomGuy

Original post by noodlemon

yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?

Can 1 + x^4 be negative?

Reply 6

noodlemon

Original post by ThatRandomGuy

Can 1 + x^4 be negative?

nope, as x^4 is always positive, plus one therefore will be +ve too? however if you put that to the power of (3/2) im failing to realize why that's always +ve, ie, why is x^(3/2) always >0? as isnt that just square root (x^3), which surely is a plus or minus? im really confused here :s

Reply 7

aznkid66

Original post by noodlemon

yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?

What makes you think this? Do you see a plus-or-minus sign in front of the root?

You only get positive and negative square roots when, for example, you use algebra to undo a squaring.

Reply 8

Indeterminate

Original post by noodlemon

yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?

Well, what goes wrong when you try to solve

\dfrac{x^2(3+x^4)}{(1+x^4)^{1.5}} < 0

Reply 9

ThatRandomGuy

Original post by noodlemon

If 1 + x^4 is always positive then it's cube will always be positive which means that it's square root will also always be positive.

Reply 10

noodlemon

Original post by ThatRandomGuy

If 1 + x^4 is always positive then it's cube will always be positive which means that it's square root will also always be positive.

ooh im sorry, im not deliberately trying to be dumb, or difficult. But say lets take (1+x^4)^3 to equal x, and we know for a fact now that x>0 , as stated before. Now we try to root this, surely a root of a +ve number, can be positive or negative? say we take x=9, surely root x is +-3?

Reply 11

noodlemon

Original post by Indeterminate

Well, what goes wrong when you try to solve

\dfrac{x^2(3+x^4)}{(1+x^4)^{1.5}} < 0

ah i see that, thank you, thats effectively proven it, but could you please read my lsat post which i quoted thatrandomguy. but thanks, i accept that

Reply 12

james22

Original post by noodlemon

The square root of a number is definded as the positive root. If you raise a number to a power e.g. x^1.5 this can only be 1 number, not 2.

Reply 13

aznkid66

Original post by noodlemon

ah i see that, thank you, thats effectively proven it, but could you please read my lsat post which i quoted thatrandomguy. but thanks, i accept that

I already answered your question: the thing is that because you differentiated the square root sign here is a positive square root sign (take the positive solution only). The fact that two values x=±k satisfy x^2=k^2 is irrelevant here.

Reply 14

ThatRandomGuy

Original post by noodlemon

No, I think you are getting confused.

If x^2 = 9 then x could equal +-3. However if x = 9 then root x will only be positive 3.

Reply 15

TenOfThem

Original post by noodlemon

thatrandomguy's post was confusing tbh

The square root of a number is defined as the +ve square root

Reply 16

noodlemon

Original post by TenOfThem

thatrandomguy's post was confusing tbh

The square root of a number is defined as the +ve square root

Oh i see... wow ive never thought of it that way, sorry for slow reply, i was getting my head round it. So effectively ive defined it to be positive already :P