Let y=cos x+sin x/cos x -sin x. Find dy/dx
if you mean
y=(cos x+sin x)/(cos x -sin x)
y' = 2/(cosx - sinx)^2
using the quotient rule, you get a bunch of terms cancelling in the numerator.
To find dy/dx, we need to differentiate the expression y = (cos x + sin x) / (cos x - sin x) with respect to x.
Let's use the quotient rule, which states that if we have a function f(x) = g(x) / h(x), then the derivative of f(x) with respect to x can be computed as:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
In our case, g(x) = cos x + sin x and h(x) = cos x - sin x. Let's find their derivatives:
g'(x) = (-sin x + cos x) (using the derivative of cos x (=-sin x) and the derivative of sin x (=cos x))
h'(x) = (-sin x - cos x) (using the derivative of cos x (=-sin x) and the derivative of sin x (=cos x))
Now, let's substitute these values into the quotient rule formula and simplify:
dy/dx = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= ((-sin x + cos x) * (cos x - sin x) - (cos x + sin x) * (-sin x - cos x)) / (cos x - sin x)^2
= (-(sin x * cos x) + cos^2 x - sin^2 x + sin x * cos x + sin x * cos x + sin^2 x + cos^2 x) / (cos x - sin x)^2
= 2cos^2 x / (cos x - sin x)^2
Therefore, dy/dx = 2cos^2 x / (cos x - sin x)^2.