Could some please help me with this, show me how to do them, I've been trying for 2 hours now and I always get the wrong answer.
y = ((4x) / (7x - 3))^1/4 find dy/dx
And, determine the slope of the following curve at the given point:
6xy^1/3 + 5y = 35; (5,1)
If I can just learn how to do these two I can do the rest of the assignment, I'm just having tons trouble figuring it out. Thanks a bunch for any help
y = ((4x) / (7x - 3))^1/4 find dy/dx
And, determine the slope of the following curve at the given point:
6xy^1/3 + 5y = 35; (5,1)
If I can just learn how to do these two I can do the rest of the assignment, I'm just having tons trouble figuring it out. Thanks a bunch for any help
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y = ((4x) / (7x - 3))^1/4
y = 4x (7x-3)^(-1/4)
Apply the product rule
dy/dx = 4 (7x-3)^(-1/4) + 4x (-1/4) (7x-3)^(-1/4-1) d/dx (7x-3)
dy/dx = 4(7x-3)^(-1/4) + (4x)(-1/4) (7x-3)^(-5/4) (7)
dy/dx = 4 / (7x-3)^(1/4) - 7x / (7x-3)^(5/4)
--------------------------------------…
6xy^1/3 + 5y = 35
differentiate both sides with respect to x
6 y^(1/3) + 6x (1/3) y^(1/3-1) dy/dx + 5 dy/dx =0
6 y^(1/3) + 2x y^(-2/3) dy/dx + 5 dy/dx = 0
dy/dx [ 2x /y^(2/3) +5] = -6y^(1/3)
dy/dx = -6y^(1/3) / [2x /y^(2/3) +5]
substitute x=5, y=1 into dy/dx
-6 (1) / [10/1+5]
= -6/15
= -2/5
y = 4x (7x-3)^(-1/4)
Apply the product rule
dy/dx = 4 (7x-3)^(-1/4) + 4x (-1/4) (7x-3)^(-1/4-1) d/dx (7x-3)
dy/dx = 4(7x-3)^(-1/4) + (4x)(-1/4) (7x-3)^(-5/4) (7)
dy/dx = 4 / (7x-3)^(1/4) - 7x / (7x-3)^(5/4)
--------------------------------------…
6xy^1/3 + 5y = 35
differentiate both sides with respect to x
6 y^(1/3) + 6x (1/3) y^(1/3-1) dy/dx + 5 dy/dx =0
6 y^(1/3) + 2x y^(-2/3) dy/dx + 5 dy/dx = 0
dy/dx [ 2x /y^(2/3) +5] = -6y^(1/3)
dy/dx = -6y^(1/3) / [2x /y^(2/3) +5]
substitute x=5, y=1 into dy/dx
-6 (1) / [10/1+5]
= -6/15
= -2/5