Help me i have big problem with these derivatives?

Question

Determine the derivative of f(x) = 125 / (6x^2 + 5x +7)^1/3 Please show your work it will help me understand

Answer 1

No problem, i just finished a calc3 course in college.

Your problem is the same as 125 * (6x^2 + 5x + 7)^(-1/3)
You have to use the chain rule so you get
125 * (-1/3) * (12x + 5) * (6x^2 +5x + 7)^(-4/3)

then you can simplify how you like from there

Answer 2

You have to use the chain rule.

Think of it this way -- let y = 6x^2+5x+7

Then we could look at f(y)

f(y) = 125/y^(1/3)

But we can rewrite this as:

f(y) = 125*y^(-1/3)

We can differentiate this using the x^n rule

f'(y) = (-1/3)*125*y^(-1/3-1) = -(125/3)*y^(-4.3)

But we don't want the derivative with respect to 'y' -- we want it with respect to 'x'

In order to do this, we replace y with 6x^2+5x+7 and apply the chain rule. The chain rule says to multiply the formula we already have by the derivative of 6x^2+5x+7 -- which is (12x+5)

Answer 3

Ok....do it stelp by step and don't let it overwhelm you.

What's 125 / (6x^2 + 5x +7)^1/3 ?

==> 125(6x^2 + 5x +7)^(-1/3)

Now everything's in the numerator......

......then:

(-1/3)*125(6x^2 + 5x +7)^(-4/3)
==> [(-125/3)(6x^2 + 5x +7)^(-4/3)](12x + 5)
==> (-125/3)(12x + 5)/((6x^2 + 5x +7)^(4/3)

Hope this sheds some light on it for you.

Answer 4

ok, here's how you do it

you declare U and dU

U= 6x^2 + 5x +7

and

dU= 12x+5

now you substitute U

f(x)=125/U^1/3

now you pass the U^1/3 besides the 125

f(U)=125*U^-1/3

And you derive

f'(U)=(-125/3)*U^-4/3dU

so finally you substitute

f'(x)=(-125/3)*(6x^2 + 5x +7)^-4/3*(12x+5)

you rearrange the equation and you get

f'(x)=(-125/3)*(12x+5)/(6x^2 + 5x +7)^4/3

hope this helps.

Answer 5

f(x) = 125 * (6x² + 5x + 7)^(-1/3)
ok
u = 5x²+5x+7
f(x) = 125 u^(-1/3)
f'(x) = -(1/3)*125 du/dx u^(-4/3)
du/dx = 12x+5
f'(x) = -125(12x+5) / 3(6x² + 5x + 7)^(4/3)