Find the differential dy of the given function.
y=sec^2x/(x^2+1)
My answer was [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2
My friend (who's a lot smarter than me) got something wayyy different, so I'm not sure if I'm doing this wrong...?
[(x^2+1)d(sec^2x)-sec^2xd(x^2+1)]/(x^2+1)^2
[(x^2+1)2secxdsecx-sec^2x(2xdx)]/(x^2+1)^2
[2(x^2+1)sec^2xtanxdx-2xsec^2xdx]/(x+1)^2
2 sec^2 x dx [(x^2+1)tan x - x ]/ (x+1)^2
Wait, I'm not understanding how you get the tanx -x part...
Hold on...I think I just managed to kind of figure it out!
To find the differential dy of the given function, y = sec^2x/(x^2+1), we can start by using the quotient rule of differentiation.
The quotient rule states that if we have a function f(x) = g(x)/h(x), then its derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Now, let's apply the quotient rule to our function y = sec^2x/(x^2+1).
We can rewrite the function as y = (sec^2x) * (x^2+1)^(-1).
Using the chain rule, the derivative of sec^2x is 2sec^2x * tan(x).
Now, applying the quotient rule, we have:
dy/dx = (2sec^2x * tan(x) * (x^2+1)^(-1) - sec^2x * (-2x)) / (x^2+1)^2
Simplifying further, we can write the derivative as:
dy/dx = [2sec^2x * tan(x) * (x^2+1) + 2x * sec^2x] / (x^2+1)^2
Therefore, the correct answer for the differential dy of the given function is:
dy = [2sec^2x * tan(x) * (x^2+1) + 2x * sec^2x] dx / (x^2+1)^2
It's possible that your friend made a different assumption or made an error during the calculation. It's always a good idea to double-check the steps and calculations to ensure accuracy.