If f(x) = x/(4x + 5)^1/6, then f '(x) = ???

Answer 1

Using the Quotient rule:
[(4x+5)^1/6 (x)' - (x) [(4x +5)^1/6]']/[(4x +5)^1/6]^2
= [(4x+5)^1/6 - (1/6)(x)(4x +5)^-5/6 (4x+5)']/(4x +5)^2/6
= [(4x+5)^1/6 - (1/6)(x)(4x +5)^-5/6 (4)]/(4x +5)^2/6
= [(4x+5)^1/6 - (4/6)(x)(4x +5)^-5/6]/(4x +5)^2/6
= [(4x+5)^1/6 - (2/3)(x)(4x +5)^-5/6 (4)]/(4x +5)^2/6
Now factor out (1/3)(4x+5)^-5/6
= ((1/3)(4x +5)^-5/6) [3(4x+5)^1 - 2(x)]/(4x+5)^2/6
Move 3(4x+5)^-5/6 to the denominator and combine like terms on top
= [12x + 15 - 2x]/[3(4x +5)^5/6 (4x +5)^2/6
= [10x -15]/[3(4x+5)^7/6]

Answer 2

The general answer of f(x)^a where a is a constant is as follows:
a*f(x)^(a-1)*f'(x)
we can re-write your equation in the form:
x^(1/6) * (4x+5)^(-1/6)
In this form, its just a function of the group product where to take the derivative of this:
g(x)*f(x)
g'(x)f(x) + g(x)f'(x)

So the final answer is:
1/6*x^(-5/6)(4x+5)^(-1/6) + x^(1/6) * (-1/6) (4x+5)^(-7/6) (4)

Answer 3

Multiplication rule:


If h(x) = f(x)*g(x),

the h'(x) = f(x) * g'(x) + f'(x) * g(x)


rewrite the function as
h(x) = x/(4x + 5)^1/6 = x * (4x + 5)^ (-1/6)

Here, f(x) = x and g(x) =(4x + 5)^ (-1/6)

h'(x) = x * (-1/6)*(4)*(4x + 5)^(-7/6) + 1 * (4x + 5)^ (-1/6)

= (-2x/3)(4x + 5)^(-7/6) + (4x + 5)^ (-1/6)

Answer 4

The answer is as follows
1/6{x^(-5/6)(4x+5)^(-1/6)} + x^(1/6) { (-1/6) (4x+5)^(-7/6) (4) }

Answer 5

I see what the little darn thing is about. You have 2 solutions, one more practical than the other. Since this is an invective function you can draw the hyperbola that belongs to it in a graph with axes 0x and 0y. The f'(x) is the complementary and opposite hyperbola to the first one. hope you understand me cause I don't have a great deal of experience with math-English:) The second method is by doing the math but that can be rather dangerous since not all Invective functions have an answer in Real numbers, and maybe the answer you're looking for is with the Complex numbers, so that is another fish to fry. Good luck! Don't have the time to do this function now, still if you e-mail me in 2 days about it, I think I'll resolve it.

Answer 6

f (x) read "f" of "x" needs a value. Say....f(x) is 3....now you plug 3 into where X is and then work out the equation. Eg: f(x)=3(X) + 6=0. Let f(x) = 6. Now plug it in: 3(6) + 6 = 0. Therefore 18 + 6 = 0. Now subtract 6 from both sides. You get 18 = a negative 6. This is not possible. So this equation has "no solution".

****Okay, appearantly after reading other answars, I should definitely steer away from math questions. Sorry***

Answer 7

f(x) = x / (4x + 5)^1/6
= (x)(4x + 5)^ -1/6

Use the formula: dy/dx = u dv/dx + v du/dx

Let u = x, and v = (4x + 5)^ -1/6
du/dx = 1 and dv/dx = (-4/6)(4x + 5)^ -7/6

Now place values into formula:

dy/dx = (x)[(-4/6)(4x + 5)^-7/6] + (4x + 5)^-1/6(1)
= (-2/3)(x)(4x + 5)^ -7/6 + (4x + 5)^ -1/6

Answer 8

this is of the form u/v where u and v are functions of x.

If y = u/v, then y' = (vu' - uv')/v^2

Here make x as u and (4x+5)^1/6 as v and find y'

Answer 9

Yes, figured it out: the answer is chocolate.