Proving that both sides are equal.

[YehiaMedhat]

[math]f'(x)=\frac {x}{cos(t)}\\ f"(x)=\frac {1}{sin(t)cos(t)^2}[/math]I tried to substitute in each choice but no use. for instance the first one[math]\frac {cos(t)}{sin(t)cos(t)} - \frac {sin(t)^2}{cos(t)}\\ \frac {1-sin(t)^3}{sin(t)cos(t)}[/math]

 

[Dr.Peterson]

[math]f'(x)=\frac {x}{cos(t)}\\ f"(x)=\frac {1}{sin(t)cos(t)^2}[/math]I tried to substitute in each choice but no use. for instance the first one[math]\frac {cos(t)}{sin(t)cos(t)} - \frac {sin(t)^2}{cos(t)}\\ \frac {1-sin(t)^3}{sin(t)cos(t)}[/math]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 [imath]x[/imath], [imath]\frac{dy}{dx}[/imath], and [imath]\frac{d^2y}{dx^2}[/imath] all in terms of [imath]t[/imath] 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.

 

[YehiaMedhat]

?, 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.

 

[topsquark]

?, 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.

Don't worry about it. Someone else might benefit from seeing it.

-Dan