Math Help - derivitive

1. derivitive

take derivitive:

y=arcsin(2t)^(1/2)

Moderator edit: Also asked here: http://www.mathhelpforum.com/math-he...erivative.html

2. Use the derivative of the inverse sine function:

$\frac{d(sin^{-1}x)}{dx}=\frac{1}{\sqrt{1-x^{2}}}$;

as well as the chain rule.

3. Can you show a step by step solution.

4. Theres really only two steps here. You're just plug and chugging. What is X in this case? What is The derivative of X. What does the chain rule say to do. Answer those questions first, and if your answers are incorrect I will do the steps.

5. the x is sqrt(2t), and the chain rule is the derivative multiplied by the inside multplied by the derivative of the inside

6. the derivative of x is 1/sqrt(2t)

7. Ok awesome. So now multiply the derivative of the "X" you just calculated, by the derivative of the sine inverse function, making sure to substitute your "X" into the portion marked x-squared.

8. What is the end solution, so that we can compare?

9. What did you get and I can let you know if you did it right or not.

10. 1/(sqrt(2)sqrt(t-2t^2))

11. Hmm. Not quite.

$\frac{d(sin^{-1}\sqrt{2t})}{dx}=\frac{2}{2\sqrt{2t}}*\frac{1}{\s qrt{1-(\sqrt{2t})^{2}}}$
$\frac{d(sin^{-1}\sqrt{2t})}{dx}=\frac{1}{\sqrt{2t}\sqrt{1-2t}}$

Remember that:

$(\sqrt{x})^{2}=x$

12. I have the same problem. The answer that you gave is not the answer that came from our book. And I cannot make the connection between the two. The answer given was sqrt2/sqrt(1-2t^2).

13. Originally Posted by 10colbathb
I have the same problem. The answer that you gave is not the answer that came from our book. And I cannot make the connection between the two. The answer given was sqrt2/sqrt(1-2t^2).
The answer in the book is not possible.

EDIT: Please check. Is the problem:

$\frac{d(sin^{-1}(\sqrt{2t})}{dx}$

Or -

$\frac{d(sin^{-1}(\sqrt{2}*t)}{dx}$

14. The problem originally stated is the correct one.

15. Originally Posted by 10colbathb
The problem originally stated is the correct one.
There is an error in your book then. In the radicand of our derivative:

$(\sqrt{2t})^2\neq2t^{2}$

And the derivative of the argument inside the sine inverse, does not produce a $\sqrt{2}$ in the numerator, even after rationalizing the $\sqrt{2t}$ in the denominator of both or our problems.