use the chain rule, in leibniz notation, to find dy/dx at the indicated value of x.
y = u^3 - 2(u - 3)^4, u = x + x^1/3, x = 1
dy/dx = dy/du du/dx
= (3u^2 - 8(u-3)^3 ) (1 + 1/3 x^-2/3)
u(1) = 2 so at x=2,
dy/dx = (3*2^2 - 8(1)^3) (1 + 1/3) = 16/3
sorry, how exactly did you get that answer? i'm confused with the steps i need to take
To find dy/dx using the chain rule in Leibniz notation, we need to differentiate y with respect to u and u with respect to x, and then multiply the two derivatives together. Let's go step by step:
Step 1: Find du/dx
Given u = x + x^(1/3), we differentiate u with respect to x:
du/dx = d/dx (x + x^(1/3))
To find the derivative of the expression x + x^(1/3), we treat x as a variable and x^(1/3) as a function of x. Therefore, we have:
du/dx = 1 + (1/3)x^(-2/3)
Step 2: Find dy/du
Given y = u^3 - 2(u - 3)^4, we differentiate y with respect to u:
dy/du = d/dx (u^3 - 2(u - 3)^4)
To differentiate y with respect to u, we treat u as a variable and expand the expression:
dy/du = 3u^2 - 8(u - 3)^3
Step 3: Find dy/dx
To find dy/dx, we multiply du/dx and dy/du:
dy/dx = (dy/du) * (du/dx)
Substituting the derivatives we found from Step 1 and Step 2:
dy/dx = (3u^2 - 8(u - 3)^3) * (1 + (1/3)x^(-2/3))
Step 4: Evaluate dy/dx at x = 1
Finally, we evaluate dy/dx at x = 1. Substituting x = 1 and u = x + x^(1/3) into the expression:
u = 1 + 1^(1/3) = 1 + 1 = 2
dy/dx = (3(2)^2 - 8(2 - 3)^3) * (1 + (1/3)(1)^(-2/3))
Simplifying further:
dy/dx = (3(4) - 8(-1)^3) * (1 + (1/3)(1)^(-2/3))
= (12 + 8) * (1 + 1/3)
= 20 * 4/3
= 80/3
Therefore, dy/dx at x = 1 is 80/3.