Proving that both sides are equal.

YehiaMedhat

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f(x)=xcos(t)f"(x)=1sin(t)cos(t)2f'(x)=\frac {x}{cos(t)}\\ f"(x)=\frac {1}{sin(t)cos(t)^2}I tried to substitute in each choice but no use. for instance the first onecos(t)sin(t)cos(t)sin(t)2cos(t)1sin(t)3sin(t)cos(t)\frac {cos(t)}{sin(t)cos(t)} - \frac {sin(t)^2}{cos(t)}\\ \frac {1-sin(t)^3}{sin(t)cos(t)}Screenshot_20221217-144306_Xodo Docs.jpg
 
f(x)=xcos(t)f"(x)=1sin(t)cos(t)2f'(x)=\frac {x}{cos(t)}\\ f"(x)=\frac {1}{sin(t)cos(t)^2}I tried to substitute in each choice but no use. for instance the first onecos(t)sin(t)cos(t)sin(t)2cos(t)1sin(t)3sin(t)cos(t)\frac {cos(t)}{sin(t)cos(t)} - \frac {sin(t)^2}{cos(t)}\\ \frac {1-sin(t)^3}{sin(t)cos(t)}View attachment 34655
What is "f"? It isn't defined in the problem. Are you assuming y = f(x), or something else?

I would express xx, dydx\frac{dy}{dx}, and d2ydx2\frac{d^2y}{dx^2} all in terms of tt alone, before trying the results in the equations. I found that one works. And I got something different from you for the second derivative, so you may want to show us your work to obtain that.
 
?, yes that's a breaking mistake with the second derivative!
Can i just delete the thread it's silly to let others read my question.
 
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