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Find the derivative of 

y = t/(sqrt(ct^2-1))

 Jun 5, 2014

Best Answer 

 #1
avatar+26367 
+6

The derivate of  $$y=\dfrac{t}{\sqrt{(ct^2-1)}} = t*\left( ct^2-1\right)^{-\frac{1}{2}}$$

Product Rule: $$y^\prime=t * \left( \frac{d(ct^2-1)^{-\frac{1}{2}}}{dt} \right)+\frac{d(t)}{dt}*(ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=t *(-\frac{1}{\not{2}})(ct^2-1)^{-\frac{1}{2}-1}}*\not{2}ct + 1*(ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=-ct^2(ct^2-1)^{-\frac{3}{2}}} + (ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=\dfrac{1}{\sqrt{(ct^2-1)}} -\dfrac{ct^2}{
\left(
\sqrt{(ct^2-1)}
\right)^3
}$$

$$\boxed{y^\prime=\dfrac{1}{\sqrt{(ct^2-1)}}
\left(
1-\dfrac{ct^2}{ct^2-1}
\right)}$$

 Jun 5, 2014
 #1
avatar+26367 
+6
Best Answer

The derivate of  $$y=\dfrac{t}{\sqrt{(ct^2-1)}} = t*\left( ct^2-1\right)^{-\frac{1}{2}}$$

Product Rule: $$y^\prime=t * \left( \frac{d(ct^2-1)^{-\frac{1}{2}}}{dt} \right)+\frac{d(t)}{dt}*(ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=t *(-\frac{1}{\not{2}})(ct^2-1)^{-\frac{1}{2}-1}}*\not{2}ct + 1*(ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=-ct^2(ct^2-1)^{-\frac{3}{2}}} + (ct^2-1)^{-\frac{1}{2}}$$

$$y^\prime=\dfrac{1}{\sqrt{(ct^2-1)}} -\dfrac{ct^2}{
\left(
\sqrt{(ct^2-1)}
\right)^3
}$$

$$\boxed{y^\prime=\dfrac{1}{\sqrt{(ct^2-1)}}
\left(
1-\dfrac{ct^2}{ct^2-1}
\right)}$$

heureka Jun 5, 2014
 #2
avatar+25 
0

Thank you Heureka 

 Jun 5, 2014

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