Quotient Problem

[Vulcan]

Working through the following problem and not sure how to get from my last step to the given solution. Any pointers please!

 

[stapel]

Working through the following problem and not sure how to get from my last step to the given solution. Any pointers please!

. . .\(\displaystyle \mbox{Differentiate }\, y\, =\, \dfrac{3x^2\, +\, 2x\, +\, 1}{1\, +\, e^x}\)

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{(6x\, +\, 2)\, (1\, +\, e^x)\, -\, (3x^2\, +\, 2x\, +\, 1)\, e^x}{(1\, +\, e^x)^2}\)

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{6x\, +\, 2}{1\, +\, e^x}\, -\, \dfrac{e^x\, (3x^2\, +\, 2x\, +\, 1)}{(1\, +\, e^x)^2}\)

Their answer:

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, -\, \dfrac{(3x^2\, -\, 4x\, -\, 1)\, e^x\, -\, 6x\, -\, 2}{(1\, +\, e^x)^2}\)

Instead of splitting the one fraction into two, try first simplifying across the top:

. . . . .\(\displaystyle (6x\, +\, 2)\, (1\, +\, e^x)\, -\, (3x^2\, +\, 2x\, +\, 1)\, e^x\)

. . . . .\(\displaystyle 6x\, +\, 2\, +\, (6x\, +\, 2)\, e^x\, -\, (3x^2\, +\, 2x\, +\, 1)\, e^x\)

. . . . .\(\displaystyle 6x\, +\, 2\, +\, (6x\, +\, 2\, -\, 3x^2\, -\, 2x\, -\, 1)\, e^x\)

. . . . .\(\displaystyle 6x\, +\, 2\, +\, (-3x^2\, +\, 4x\, +\, 1)\, e^x\)

. . . . .\(\displaystyle -\, \bigg(-6x\, -\, 2\, +\, (3x^2\, -\, 4x\, -\, 1)\, e^x\, \bigg)\)

Granted, the above isn't how I would do the simplification but, as you can see, your form and theirs are equivalent.

 

[Steven G]

Working through the following problem and not sure how to get from my last step to the given solution. Any pointers please!

Whenever part of your answer matches the answer you want do not EVER give that up. You had the denominator you want-perfect! So all that is left is to get the numerator to become what you want.

On another note, you want your final answer to have a common denominator so when you have one don't break it up.

 

[Vulcan]

Instead of splitting the one fraction into two, try first simplifying across the top:

. . . . .\(\displaystyle (6x\, +\, 2)\, (1\, +\, e^x)\, -\, (3x^2\, +\, 2x\, +\, 1)\, e^x\)

. . . . .\(\displaystyle 6x\, +\, 2\, +\, (6x\, +\, 2)\, e^x\, -\, (3x^2\, +\, 2x\, +\, 1)\, e^x\)

. . . . .\(\displaystyle 6x\, +\, 2\, +\, (6x\, +\, 2\, -\, 3x^2\, -\, 2x\, -\, 1)\, e^x\)

. . . . .\(\displaystyle 6x\, +\, 2\, +\, (-3x^2\, +\, 4x\, +\, 1)\, e^x\)

. . . . .\(\displaystyle -\, \bigg(-6x\, -\, 2\, +\, (3x^2\, -\, 4x\, -\, 1)\, e^x\, \bigg)\)

Granted, the above isn't how I would do the simplification but, as you can see, your form and theirs are equivalent.

I had another go at it last night and came up with the following! It's maybe a bit long winded but it looks right!