y= cube root (1+tan(t)) OR
y= (1+tan(t))^(1/3)
The answer I got was
y'= (1/3)(1+tan(t))^(-2/3)*(sec^2(t))
Is this correct?
I had to find dy/dx or y' same difference. I just want to make sure i am doing it right because i have a hard time with trig functions
Correct!
To check if the derivative you obtained is correct, let's go through the steps of finding the derivative of the function.
We have the function: y = (1 + tan(t))^(1/3)
To find the derivative of y with respect to t (dy/dt), we can use the chain rule.
1. Start by rewriting the function as y = (1 + tan(t))^((1/3)*(1)), where we notice that (1/3) is the exponent.
2. Apply the chain rule:
a. Take the derivative of the outer function, which is the power function: d/dt(u^n) = n*u^(n-1)*(du/dt), where u = 1 + tan(t) and n = (1/3).
Therefore, d/dt[(1 + tan(t))^((1/3)*(1))] = (1/3)*(1 + tan(t))^((1/3)*(1) - 1)*(d/dt(1 + tan(t))).
b. Now, find the derivative of the inner function, which is (1 + tan(t)): d/dt(1 + tan(t)) = sec^2(t).
3. Substitute the values we found back into the chain rule:
dy/dt = (1/3)*(1 + tan(t))^((1/3)*(1) - 1)*(sec^2(t)).
Simplifying the exponent:
(1/3)*(1/3)*(1) - 1 = (1/3) - 1 = (1/3) - (3/3) = -2/3.
Substituting the simplified exponent into the equation:
dy/dt = (1/3)*(1 + tan(t))^(-2/3)*(sec^2(t)).
So, after going through the steps, we can see that the derivative you obtained:
y' = (1/3)*(1 + tan(t))^(-2/3)*(sec^2(t))
is indeed correct.