how to find derivative of y:
y = sin inverse(square root of 2t)
I got this. Please ignore.
Finding the derivative of a function involves using the rules of calculus. In this case, you want to find the derivative of y = sin^(-1)(√(2t)).
To find the derivative, you can use the chain rule. The chain rule states that if you have a composite function, such as f(g(x)), then the derivative of f(g(x)) is equal to the derivative of f with respect to g, multiplied by the derivative of g with respect to x.
In your case, let f(u) = sin^(-1)(u) and g(t) = √(2t). We want to find the derivative of y = f(g(t)).
Step 1: Find the derivative of f(u) = sin^(-1)(u).
The derivative of sin^(-1)(u) with respect to u is 1/√(1-u^2).
Step 2: Find the derivative of g(t) = √(2t).
The derivative of √(2t) with respect to t is (1/2√(2t)) * 2 = 1/√(2t).
Step 3: Apply the chain rule.
The derivative of y = f(g(t)) is (1/√(1-(√(2t))^2)) * (1/√(2t)).
Simplifying further, we have:
(1/√(1-2t))*(1/√(2t)).
Combining the square roots in the denominator:
(1/√((1-2t)*2t)).
So, the derivative of y = sin^(-1)(√(2t)) is (1/√((1-2t)*2t)).
Remember to always double-check your work and consult with a math expert if needed.