Hi guys, i attached the question with the last answear that i got (which i think is right) . I have attempted this question more then 20 times, and ready to give up. Please have a look, and see what you think
Hi guys, i attached the question with the last answear that i got (which i think is right) . I have attempted this question more then 20 times, and ready to give up. Please have a look, and see what you think
That answer would be right if the function was $\displaystyle \sqrt{1-121x^2}\arcsin(11x)$
Just take in mind that
$\displaystyle \frac{d}{du} \arccos u= -\frac{1}{\sqrt{1-u^2}}$
The answer you should get is very similar to the one you got before.
It would help if you show your work step by step so we can detect where's the mistake.
Hi iFuuZe.
To calculate $\displaystyle g'(x)$ when $\displaystyle g(x)=\sqrt{1-121x^2}\cdot cos^{-1}(11x)$ you would start by assuming that you need to use the product rule, right? But hold on.
$\displaystyle \sqrt{1-121x^2}$ can be rewritten as $\displaystyle (1-121x^2)^{\frac{1}{2}}$ so we will actually start by applying the chain rule.
$\displaystyle \frac{d}{dx}(1-121x^2)^{\frac{1}{2}}=\frac{1}{2}\left(1-121x^2\right)^{-\frac{1}{2}}\left(-242x\right)$ all while applying the product rule
$\displaystyle \frac{d}{dx}(1-121x^2)^{\frac{1}{2}}=\frac{1}{2}\left(1-121x^2\right)^{-\frac{1}{2}}\left(-242x\right)\cdot cos^{-1}(11x)+\left(1-121x^2\right)^{\frac{1}{2}}\left(\frac{-11}{\sqrt{1-121x^2}}\right)$
With some algebraic manipulation you will end up with
$\displaystyle \frac{-121x\cdot cos^{-1}(11x)}{\sqrt{1-121x^2}}-11$ which should be the right answer. Please let me know if you are having problems understanding any of the steps.