y= u+3/u-1 u=x^1/3 find dy/dx at x=8
dy/dx = dy/du du/dx
du/dx = (1/3)x^-(2/3)
dy/du = [(u-1)-(u+3)]/(u-1)^2
= [(x^(1/3)-1)-(x^(1/3)+3)/(x^(1/3)-1)^2
= -4 / (x^(2/3) -2x^(1/3)+1)
so
dy/du * du/dx
=(-4/3)(x^-(2/3)) /(x^(2/3) -2x^(1/3)+1)
= -4/3(x^4/3-2x +x^(2/3))
luckily 8^(1/3) = 2 :)
To find dy/dx, we need to differentiate the equation y = u + 3/u - 1 with respect to x.
First, let's rewrite the equation using the given expression for u: u = x^(1/3).
Substituting this value of u into the equation, we have:
y = x^(1/3) + 3/(x^(1/3)) - 1.
Next, we differentiate both sides of this equation with respect to x, using the rules of differentiation.
The derivative of the first term, x^(1/3), can be found using the power rule. Since the power is 1/3, we multiply by the reciprocal of the power and reduce it by 1:
d/dx (x^(1/3)) = (1/3) * x^(-2/3) = 1/(3x^(2/3)).
Now, let's find the derivative of the second term, 3/(x^(1/3)).
To differentiate this term, we can rewrite it as 3 * x^(-1/3). Similar to the previous term, we use the power rule:
d/dx (3 * x^(-1/3)) = -1/3 * 3 * x^(-4/3) = -x^(-4/3)/x^(-1/3) = -x^(-4/3 + 1/3) = -x^(-1).
Finally, the derivative of the constant term -1 is 0.
Adding up the derivatives of all three terms, we have:
dy/dx = 1/(3x^(2/3)) - x^(-1) + 0.
Now, we can determine the value of dy/dx at x = 8 by substituting x = 8 into the expression we obtained:
dy/dx = 1/(3(8^(2/3))) - 8^(-1).
Simplifying further,
dy/dx = 1/(24∛8) - 1/8 = 1/(24 * 2) - 1/8 = 1/48 - 1/8 = (1 - 6)/48 = -5/48.
Therefore, dy/dx at x = 8 is -5/48.