Find the derivative of the following questions;
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
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
where did you get u from
To find the derivative of the function ³√(2t) / (t^2-4), we can use the quotient rule. The quotient rule states that if we have a function in the form f(x) = g(x) / h(x), then the derivative is given by the formula:
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.