Asked by Alex
Let y = u^3+6u, where u = 2x^4+3x^2-2
Find dy/dx.
Also..
2) Let f(x) = 1/sqrt(4x^2-x). Determine f'(x).
3) Let y = 5/cubrt((2x-5)^2). Determine dy/dx.
Find dy/dx.
Also..
2) Let f(x) = 1/sqrt(4x^2-x). Determine f'(x).
3) Let y = 5/cubrt((2x-5)^2). Determine dy/dx.
Answers
Answered by
drwls
1) dy/dx = dy/du*du/dx
= (3 u^2 + 6)(8x^3 + 6x)
To get it in terms of x only, substitute the u(x) function for u.
2) Use the same "function of a function" approach as above. Let
u = 4x^2 -x
f(u) = u^(-1/2)
df/dx = df/du*du/dx
3) Let u = (2x-5)
f(u) = 5/u^(2/3)
Use the same procedure
= (3 u^2 + 6)(8x^3 + 6x)
To get it in terms of x only, substitute the u(x) function for u.
2) Use the same "function of a function" approach as above. Let
u = 4x^2 -x
f(u) = u^(-1/2)
df/dx = df/du*du/dx
3) Let u = (2x-5)
f(u) = 5/u^(2/3)
Use the same procedure
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