Differentiating t/(1+t^2)^(1/2)

In summary, the derivative of t/(1+t<sup>2</sup>)<sup>(1/2)</sup> is (1+t<sup>2</sup>)<sup>(-3/2)</sup> * (1 - t<sup>2</sup>), and the expression can be simplified using the power and chain rules. However, it cannot be simplified further and the critical point is t=0. The domain of the expression is all real numbers, except for t=±i.
  • #1
P.O.L.A.R
22
0
d/dx[t/(1+t^2)^(1/2)]

The answer say 1/(1+t^2)^(3/2)

I can tget that for the life of me. Used product rule and I can't seem to simplify it I come up with

2/(1+t^2)^(3/2) Where does the 2 go? This is for a principal unit vector problem so a quick solution is all i need. thanks
 
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  • #2
nevermind i got it forgot about the 2 on the bottom doh.
 
  • #3
I'd say the answer was 0, but that's just my opinion..
 
  • #4
arildno said:
I'd say the answer was 0, but that's just my opinion..

lol...
 

Related to Differentiating t/(1+t^2)^(1/2)

1. What is the derivative of t/(1+t2)(1/2)?

The derivative of t/(1+t2)(1/2) is (1+t2)(-3/2) * (1 - t2), also known as the quotient rule.

2. How do you simplify the expression t/(1+t2)(1/2) before finding the derivative?

You can use the power rule to simplify the expression to t(1+t2)(-1/2), then use the chain rule to find the derivative.

3. Can the expression t/(1+t2)(1/2) be simplified further?

No, t/(1+t2)(1/2) is already in its simplest form.

4. How do you find the critical points of t/(1+t2)(1/2)?

To find the critical points, set the derivative equal to 0 and solve for t. In this case, the critical point is t=0.

5. What is the domain of t/(1+t2)(1/2)?

The domain of t/(1+t2)(1/2) is all real numbers, except for t=±i, where i is the imaginary unit.

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