can someone actually show me all the steps to finding this deriviative. I keep doing the problem out and getting the wrong answer.
Find the deriviative of:
INVERSEsin(2t)^1/2
What is under the square root sign? the way you have written it, it is the argument of the sine function, 2t
Assuming that is it
y= INVsin(sqrt2t)
I am going to shift to arcsin notation, so we don't get messed up in power notation.
y=arcsin(sqrt2t)
y=arcsin u where u=sqrt2t
y'=1/(1-u^2) du/dx
and du/dt= 1/(2sqrt2t) * 2 or 1/sqrt2t
y=1/((1-2t)(sqrt2t))
To find the derivative of the function INVERSEsin(2t)^(1/2), you can use the chain rule, which states that if you have a composite function of the form f(g(t)), then the derivative of that function with respect to t is given by f'(g(t)) * g'(t).
First, let's find the derivative of the inner function g(t) = sin(2t). The derivative of sin(u) with respect to u is cos(u). Therefore, the derivative of sin(2t) with respect to t is cos(2t).
Next, we need to find the derivative of the outer function f(u) = u^(1/2). The power rule states that the derivative of u^n with respect to u is n*u^(n-1). Therefore, the derivative of u^(1/2) with respect to u is (1/2)*u^(-1/2).
Now we can combine these results using the chain rule. The derivative of INVERSEsin(2t)^(1/2) with respect to t is:
f'(g(t)) * g'(t) = (1/2)*(INVERSEsin(2t))^(-1/2) * cos(2t)
So, the derivative of INVERSEsin(2t)^(1/2) with respect to t is (1/2)*(INVERSEsin(2t))^(-1/2) * cos(2t).