find the derivative of y with respect to the appropriate variable
y=sec^-1*5s
To find the derivative of y with respect to the appropriate variable, we can use the chain rule of differentiation. The chain rule states that if we have a composition of functions, such as y = f(g(x)), then the derivative of y with respect to x is given by:
dy/dx = f'(g(x)) * g'(x)
In this case, we have y = sec^-1(5s). To apply the chain rule, we need to identify the inner function and its derivative, as well as the outer function and its derivative.
The inner function is g(x) = 5s, and its derivative is g'(x) = 5 (since the derivative of a constant multiplied by x is just the constant).
The outer function is f(u) = sec^-1(u), and to find its derivative, we need to use the chain rule again. The derivative of sec^-1(u) is given by:
f'(u) = 1 / (|u|sqrt(u^2 - 1))
Now we can apply the chain rule:
dy/dx = f'(g(x)) * g'(x)
= [1 / (|g(x)|sqrt(g(x)^2 - 1))] * g'(x)
= [1 / (|5s|sqrt((5s)^2 - 1))] * 5
Simplifying further:
dy/dx = 5 / (|5s|sqrt(25s^2 - 1))
So the derivative of y with respect to the appropriate variable (assuming it is 's') is given by 5 / (|5s|sqrt(25s^2 - 1)).