Find the dy/dx. arcsin ($\displaystyle \sqrt{x}$) I got $\displaystyle \frac{\sqrt{x}}{\sqrt{1-2t}}$ This, is not correct, I used the typical trig formula for this. 1/(1-u^2) * du/dx. Thanks!
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Originally Posted by Truthbetold Find the dy/dx. arcsin ($\displaystyle \sqrt{x}$) I got $\displaystyle \frac{\sqrt{x}}{\sqrt{1-2t}}$ This, is not correct, I used the typical trig formula for this. 1/(1-u^2) * du/dx. Thanks! now how in the world did you end up with t in your answer? $\displaystyle \frac d{dx} \arcsin u = \frac {u'}{\sqrt{1 - u^2}}$ where $\displaystyle u$ is a function of $\displaystyle x$ now, if $\displaystyle u = \sqrt{x}$, what is $\displaystyle u'$?
Originally Posted by Jhevon now how in the world did you end up with t in your answer? $\displaystyle \frac d{dx} \arcsin u = \frac {u'}{\sqrt{1 - u^2}}$ where $\displaystyle u$ is a function of $\displaystyle x$ now, if $\displaystyle u = \sqrt{x}$, what is $\displaystyle u'$? I don't know how I got t in my answer. I made a typo. Derivative of u'= $\displaystyle \frac{1}{2\sqrt{x}}$
Originally Posted by Truthbetold I don't know how I got t in my answer. I made a typo. Derivative of u'= $\displaystyle \frac{1}{2\sqrt{x}}$ ok, so now try again. what's the derivative of the original function?
$\displaystyle \frac{1}{2\sqrt{x} * \sqrt{1-x}}$.. my answer
Originally Posted by Truthbetold $\displaystyle \frac{1}{2\sqrt{x} * \sqrt{1-x}}$.. my answer that is correct. you could simplify it, but that's fine
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