finding derivative: f ' (x) = ( -56x ) / ( 2x^2 + 7 )^2

[calculusIsKillingMe]

finding derivative of: f ' (x) = ( -56x ) / ( 2x^2 + 7 )^2

f’(x)=-56x
(2x²+7)²

f″(x)=-56 (2x²+7)²-2x(2x²+7)4x
(2x²+7)⁴

=-56 (2x²+7)(2x²+7-8x²)
(2x²+7)4


I dont understand how the -2x and 4x dispaear and re apear as -8x² in the same bracket as (2x²+7..). I thought it should be
-56 (2x²+7)²-8x²(2x²+7)



Also, I dont understand how this final answer came.

f''(x)= 56 (√6*x-√7)(√6*x+√7)
(2x²+7)

 

[Deleted member 4993]

f’(x)=-56x
(2x²+7)²

f″(x)=-56 (2x²+7)²-2x(2x²+7)4x
..........................(2x²+7)⁴
=-56 (2x²+7)²(2x²+7-8x²) i dont understand how the -2x and 4x dispaear
..................(2x²+7)⁴ ..........and re apear as -8x² in the same bracket as (2x²+7..). I thought it should be
............................................-56 (2x²+7)²-8x²(2x²+7)

the numerator has two terms and both the terms contain (2x²+7) - thus it was factored out . However the exponent
of (2x²+7) in the numerator should be 1.


= 56 (√6*x-√7)(√6*x+√7) ..also, i dont understand how this final answer
.......................(2x²+7) .......... came.

however the exponent of (2x²+7) in the denominator should be 3.

the identity:

A² - b² = (A+b)(A-b)

was used to factorize the numerator further.

.

 

[calculusIsKillingMe]

What happend to the 2x²+7-8x^​2

 

[stapel]

f’(x)=-56x
(2x²+7)²

I think we're all guessing that the above means the following:

. . .\(\displaystyle \mbox{Differentiate and simplify:}\)

. . . . .\(\displaystyle f'(x)\, =\, \dfrac{-56x}{(2x^2\, +\, 7)^2}\)

Does your book commonly begin exercises with the first derivative? :shock:

f″(x)=-56 (2x²+7)²-2x(2x²+7)4x
(2x²+7)4
=-56 (2x²+7)(2x²+7-8x²) I dont understand how the -2x and 4x dispaear
(2x²+7)4 and re apear as -8x² in the same bracket as (2x²+7..). I thought it should be
-56 (2x²+7)²-8x²(2x²+7)

This multi-line mixture of math and text is quite confusing. (This is probably affecting the clarity of our responses, as well.) I am guessing that the above means the following:

. . .\(\displaystyle \mbox{Here is my working. I started with the Quotient Rule:}\)

. . . . .\(\displaystyle f''(x)\, =\, \dfrac{(-56)(2x^2\, +\, 7)^2\, -\, (-56x)(2)(2x^2\, +\, 7)(4x)}{\left((2x^2\, +\, 7)^2\right)^2}\)

. . . . . . .\(\displaystyle =\, \dfrac{-56(2x^2\,+\, 7)^2\, +\, 56(8x^2)(2x^2\, +\, 7)}{(2x^2\, +\, 7)^4}\)

. . . . . . .\(\displaystyle =\, \dfrac{(56)(-1)(2x^2\, +\, 7)(2x^2\, +\, 7)\, +\, (56)(8x^2)(2x^2\, +\, 7)}{(2x^2\, +\, 7)(2x^2\, +\, 7)^3}\)

But I'm afraid I don't understand your final result...?

I thought it should be
-56 (2x²+7)²-8x²(2x²+7)
= 56 (√6*x-√7)(√6*x+√7) Also, I dont understand how this final answer
(2x²+7) came.

From what you've posted (which, to be fair, included no instructions; we've had to guess), there is no apparent justification for this "answer". Sorry.

 

[calculusIsKillingMe]

Sorry about the mess. I cleaned it up now