y= (2t/(t^2-4)^1/3
let u = (t^2-4)
y=( 2tu)1/3
y=1/3 (2t/u)^(-2/3) ( 2/u + 2td(1/u/dt)
now consider the last term
d(1/u)/dt=d 1/(t^2-4)=-2t/(t^2-4)^2
check my work
³√2t/(t^2-4)
Every thing is under the ³√ and the / is a fraction bar.
My answer is y'= t(2t)^3
Can you check my answer please
let u = (t^2-4)
y=( 2tu)1/3
y=1/3 (2t/u)^(-2/3) ( 2/u + 2td(1/u/dt)
now consider the last term
d(1/u)/dt=d 1/(t^2-4)=-2t/(t^2-4)^2
check my work
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's apply the quotient rule to our function:
We have g(x) = ³√(2t) and h(x) = t^2 - 4.
First, let's find g'(x), which is the derivative of g(x):
g'(x) = d/dx(³√(2t))
To find the derivative of ³√(2t), we can rewrite it as (2t)^(1/3) and then apply the chain rule:
g'(x) = (1/3) * (2t)^(-2/3) * d/dx(2t)
g'(x) = (2/3) * (2t)^(-2/3) * 2
Simplifying:
g'(x) = 4 / (3 * (2t)^(2/3))
Next, let's find h'(x), which is the derivative of h(x):
h'(x) = d/dx(t^2 - 4)
Since t^2 - 4 is a simple polynomial, its derivative is straightforward:
h'(x) = 2t
Now, we can plug g'(x), h(x), g(x), and h'(x) into the quotient rule formula to find f'(x):
f'(x) = ((4 / (3 * (2t)^(2/3))) * (t^2 - 4) - ³√(2t) * 2t) / (t^2 - 4)^2
Simplifying further may involve distributing and combining like terms.
Comparing this result to your answer of y' = t(2t)^3, we can see that they are not the same. Thus, your answer is incorrect.
Please note that solving the derivative symbolically can be quite complex for this particular function. Simplifying the expression may yield a more concise result.