$\displaystyle y = x^2cscx\sqrt{x^2+1} $ its derivative is : $\displaystyle y' = 2xcscx\sqrt{x^2+1} - x^2cotxcscx\sqrt{x^2+1} + x^2cscx\frac{1}{2\sqrt{x^2+1}} $ correct?
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Not quite $\displaystyle y' = 2xcscx\sqrt{x^2+1} - \bigg(x^2cotxcscx\sqrt{x^2+1}\bigg)\bigg(\frac{2x} {2\sqrt{x^2+1}}\bigg)$ Oops nevermind didn't see that there was an x behinf the csc
Originally Posted by 11rdc11 Not quite $\displaystyle y' = 2xcscx\sqrt{x^2+1} - \bigg(x^2cotxcscx\sqrt{x^2+1}\bigg)\bigg(\frac{2x} {2\sqrt{x^2+1}}\bigg)$ hrmm what happened to the $\displaystyle + x^2cscx$ ? and the 2x where it come from? thank u
Originally Posted by 11rdc11 Not quite $\displaystyle y' = 2xcscx\sqrt{x^2+1} - \bigg(x^2cotxcscx\sqrt{x^2+1}\bigg)\bigg(\frac{2x} {2\sqrt{x^2+1}}\bigg)$ Oops nevermind didn't see that there was an x behinf the csc so is mine okay ?
$\displaystyle $$\displaystyle y' = 2xcscx\sqrt{x^2+1} - (x^2cotxcscx\sqrt{x^2+1}) + x^2cscx\frac{2x}{2\sqrt{x^2+1}}$$\displaystyle $ as my final answer?
Quick question is it $\displaystyle x^2csc(x\sqrt{x^2 -1})$ or $\displaystyle x^2(cscx)(\sqrt{x^2-1})$
Originally Posted by jvignacio so is mine okay ? $\displaystyle y' = 2xcscx\sqrt{x^2+1} - (x^2cotxcscx\sqrt{x^2+1}) + x^2cscx\frac{x}{\sqrt{x^2+1}} $ as my final answer?
Originally Posted by 11rdc11 Quick question is it $\displaystyle x^2csc(x\sqrt{x^2 -1})$ or $\displaystyle x^2(cscx)(\sqrt{x^2-1})$ its the 2nd one
Originally Posted by jvignacio $\displaystyle y' = 2xcscx\sqrt{x^2+1} - (x^2cotxcscx\sqrt{x^2+1}) + x^2cscx\frac{x}{\sqrt{x^2+1}} $ as my final answer? Yes, you are right.
Originally Posted by Chop Suey yes. thank u!!
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