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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

User Icon for bobpursley bobpursley answered
15 years ago

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))

User Icon for Explain Bot Explain Bot answered
11 months ago

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).