Finding the derivative #2:
- Thread starter Khan101
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blamocur
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3. Show work that you've already done (even if you think it's wrong), or try to explain why you're stuck.
(from https://www.freemathhelp.com/forum/threads/posting-guidelines-summary.109845/#post-422890)
(from https://www.freemathhelp.com/forum/threads/posting-guidelines-summary.109845/#post-422890)
D
Deleted member 4993
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How can you use the learnings of your first post:
Finding the derivative
Hey all, Wondering if anyone could provide working out to answer the following question: --> Derive 6x(√x+1).... NOTE: x+1 is within square root. Cheers,
www.freemathhelp.com
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
READ BEFORE POSTING
Please share your work/thoughts about this problem
Apologies for that mate, I am new to this site, or better yet any forum page. I have fixed this inconvenience (seen below), thank you for the notice.
Steven G
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On your 1st line it clearly shows the last x is outside of the square root sign.
On the 2nd line it clearly shows bot x's under the square root sign.
You come here for help but you me and the other helpers remember from line to line where the x really are. Personally, I prefer to help someone that takes some care in what they write.
On the 2nd line it clearly shows bot x's under the square root sign.
You come here for help but you me and the other helpers remember from line to line where the x really are. Personally, I prefer to help someone that takes some care in what they write.
This is true.
One way to reduce errors is to use substitutions. It is seldom if ever necessary to do so, but I have found it reduces my significant capacity to make mechanical errors.
[math] y = -4x \sqrt{1 - 2x}.\\ \text {Let } u = - 4x \implies u’= -4.\\ \text {Let } v = \sqrt{1 - 2x} = (1 - 2x)^{1/2} \implies v’ = \dfrac{1}{2} * (1 - 2x)^{-1/2} * (-2) = - (1 - 2x)^{-1/2} = - \dfrac{1}{v}.\\ y = uv \implies y’ = uv’ + u’v = (-4x) * \left (- \dfrac{1}{v} \right ) + (-4)v = \dfrac{4x}{v} - 4v = \dfrac{4x - 4v^2}{v}.\\ \therefore \ y’ = \dfrac{4x - 4(1 - 2x)}{\sqrt{1 - 2x}} = \dfrac{12x - 4}{\sqrt{1 - 2x}}. [/math]
I agree with you after far less work.
[math] y = -4x \sqrt{1 - 2x}.\\ \text {Let } u = - 4x \implies u’= -4.\\ \text {Let } v = \sqrt{1 - 2x} = (1 - 2x)^{1/2} \implies v’ = \dfrac{1}{2} * (1 - 2x)^{-1/2} * (-2) = - (1 - 2x)^{-1/2} = - \dfrac{1}{v}.\\ y = uv \implies y’ = uv’ + u’v = (-4x) * \left (- \dfrac{1}{v} \right ) + (-4)v = \dfrac{4x}{v} - 4v = \dfrac{4x - 4v^2}{v}.\\ \therefore \ y’ = \dfrac{4x - 4(1 - 2x)}{\sqrt{1 - 2x}} = \dfrac{12x - 4}{\sqrt{1 - 2x}}. [/math]
I agree with you after far less work.
Thank you Jeff, your help is greatly appreciated!