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Differentiation. Need guidance
- Thread starter Henrik
- Start date
D
Deleted member 4993
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Given that y=√√(1+√x)-1
show that dy/dx= 1/(8√x(1+√x)(1+√x)-1)
Can somebody explain how to do it step by step? I cant figure out where to start.
First substitute
u = 1 + √x
then apply chain rule and continue...
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/th...217#post322217
We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")
Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
HallsofIvy
Elite Member
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- Jan 27, 2012
- Messages
- 7,760
Square root of square root? Is that how it was given or was the problem 4√?Given that y=√√(1+√x)-1
show that dy/dx= 1/(8√x(1+√x)(1+√x)-1)
Can somebody explain how to do it step by step? I cant figure out where to start.
In anycase, the simplest way to do this to write it as \(\displaystyle (1+ x^{1/2})^{1/4}- 1\).
Do you know how to differentiatate \(\displaystyle x^n\)? Do you know the chain rule?
In addition to seeing your work, I need some clarification on order of operations. When typing inline, it is often necessary to use additional parentheses to avoid ambiguity. Is itGiven that y=√√(1+√x)-1
show that dy/dx= 1/(8√x(1+√x)(1+√x)-1)
Can somebody explain how to do it step by step? I cant figure out where to start.
y = √[√(1 + √x) - 1] ?
First substitute
u = 1 + √x
then apply chain rule and continue...
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/th...217#post322217
We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")
Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
What I have done:
y=(1+x^1/2)^1/2
u=1+x^1/2 y=u^1/2
u'=(x^-1/2)/2 y=(u^1/2)/2
y=((1+x^1/2)^-1/2)x^-1/2)
y=((done above^)-1)^1/2
y'=(u^-1/2)/2
I know this is wrong and the problem is the order I think. I have no clue to where i should derive first and yes I am familiar with the chain rule.
Yes.In addition to seeing your work, I need some clarification on order of operations. When typing inline, it is often necessary to use additional parentheses to avoid ambiguity. Is it
y = √[√(1 + √x) - 1] ?
And yes i am familiar with the chain rule. The problem here is that the paranthesis is confusing me. i have broken down the expression to this:
[(1+x^1/2)^1/2-1]^1/2
And tried to apply the chain rule by beginning with the inner bracket first and then the rest which was wrong.
Last edited:
D
Deleted member 4993
Guest
y = √[√(1+√x)-1]
u = √(1+√x)
du/dx = 1/2 * 1/√(1+√x) * 1/2 * 1/√x = 1/[4*√{x(1+√x)}]
y = √(u-1)
dy/du = 1/[2*√(u-1)]
dy/dx = dy/du * du/dx
continue.....
There is no need to limit the chain rule to two links. You can use substitutions and the chain rule as extensively as you want to simplify a problem. Personally, I like to make the substitutions and steps explicit because it helps me avoid mistakes.
\(\displaystyle [\alpha]\ y = \sqrt{\sqrt{1 +\sqrt{x}} - 1}.\)
\(\displaystyle [\beta]\ Let\ u = \sqrt{1 + \sqrt{x}} - 1 \implies y = \sqrt{u} = u ^{(1/2)} \implies \dfrac{dy}{du} = \dfrac{1}{2} * u^{-(1/2)} = \dfrac{1}{2\sqrt{u}}.\) Hard to make a mistake there.
\(\displaystyle [\gamma]\ Let\ v = 1 + \sqrt{x} \implies u = \sqrt{v} - 1 \implies \dfrac{du}{dv} = \dfrac{1}{2\sqrt{v}} - 0 = \dfrac{1}{2\sqrt{v}}.\) Again, very easy.
\(\displaystyle [\delta]\ And\ v = 1 + \sqrt{x} \implies \dfrac{dv}{dx} = \dfrac{1}{\sqrt{x}}.\) Not a problem in the world to do.
Now apply the chain rule:
\(\displaystyle [\epsilon]\ \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dx} = \dfrac{1}{2\sqrt{u}} * \dfrac{1}{2\sqrt{v}} * \dfrac{1}{2\sqrt{x}} = \dfrac{1}{8\sqrt{u} * \sqrt{v} * \sqrt{x}} = \dfrac{1}{8\sqrt{uvx}}.\) Again, super simple. Hard to make a mistake.
Now, using gamma, delta, and epsilon, reverse the substitutions.
\(\displaystyle \dfrac{dy}{dx} = \dfrac{1}{8\sqrt{uvx}} = \dfrac{1}{8\sqrt{x(1+ \sqrt{x})(\sqrt{v} - 1)}} = \dfrac{1}{8\sqrt{x(1 + \sqrt{x})(\sqrt{1 + x} - 1)}}.\) Not hard at all.
No step in this method of explicit substitution and chain rule is hard, but the method is lengthy and looks inelegant. As I said, it removes the mystery and prevents errors.
\(\displaystyle [\alpha]\ y = \sqrt{\sqrt{1 +\sqrt{x}} - 1}.\)
\(\displaystyle [\beta]\ Let\ u = \sqrt{1 + \sqrt{x}} - 1 \implies y = \sqrt{u} = u ^{(1/2)} \implies \dfrac{dy}{du} = \dfrac{1}{2} * u^{-(1/2)} = \dfrac{1}{2\sqrt{u}}.\) Hard to make a mistake there.
\(\displaystyle [\gamma]\ Let\ v = 1 + \sqrt{x} \implies u = \sqrt{v} - 1 \implies \dfrac{du}{dv} = \dfrac{1}{2\sqrt{v}} - 0 = \dfrac{1}{2\sqrt{v}}.\) Again, very easy.
\(\displaystyle [\delta]\ And\ v = 1 + \sqrt{x} \implies \dfrac{dv}{dx} = \dfrac{1}{\sqrt{x}}.\) Not a problem in the world to do.
Now apply the chain rule:
\(\displaystyle [\epsilon]\ \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dx} = \dfrac{1}{2\sqrt{u}} * \dfrac{1}{2\sqrt{v}} * \dfrac{1}{2\sqrt{x}} = \dfrac{1}{8\sqrt{u} * \sqrt{v} * \sqrt{x}} = \dfrac{1}{8\sqrt{uvx}}.\) Again, super simple. Hard to make a mistake.
Now, using gamma, delta, and epsilon, reverse the substitutions.
\(\displaystyle \dfrac{dy}{dx} = \dfrac{1}{8\sqrt{uvx}} = \dfrac{1}{8\sqrt{x(1+ \sqrt{x})(\sqrt{v} - 1)}} = \dfrac{1}{8\sqrt{x(1 + \sqrt{x})(\sqrt{1 + x} - 1)}}.\) Not hard at all.
No step in this method of explicit substitution and chain rule is hard, but the method is lengthy and looks inelegant. As I said, it removes the mystery and prevents errors.