00:01
In this problem, we are asked to find the derivative of 5 expressions.
00:05
Let us begin with the first expression, which is secant y added with co -secant x, which equals to log of x.
00:17
Differential of secant y is secant y tan y.
00:22
Since we are differentiating with respect to x, multiply this with d -y over d x.
00:27
Added with differential of co -secent x is negative co -secent x cotex, which equals to the differential of log x is 1 over x.
00:41
Solving for d -y over d -x we obtain 1 over secant y tan y times 1 over x added with cosecant x cortex.
00:57
So this is our final answer for the first expression.
01:15
Next, moving towards the second expression, which is y squared equals to x squared added with one.
01:23
Differenti differentiating this, we obtain 2y, dy over dx, which equals to 2x differential of a constant is 0.
01:33
2 and 2 cancel themselves.
01:35
So we obtain dy over d x.
01:38
X equals to x over y.
01:42
So this is our final answer for the second expression.
01:48
Next, moving towards the third expression which is y squared equals to four times of x squared added with x.
02:02
D this, we get 2y, dy over d x which equals to differential of x square is 2x...