Algebra problem

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tdotwire8

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Hi, I am beginning calculus right now but there's a bit of Algebra at the end of a limit problem that has me scratching my head. I'm obviously missing this knowledge as it might be very straight forward, but bear with me here.

So the limit problem is as follows F(x) = √2x

now the answer as I've checked wolframalpha is 1/(√2)(√x)

I just don't get how this is the answer. Checking the step by step calculation of this derivative I've follow the exact same until the second last step, this is where the algebra is messing with me.

the last step I get correct is the answer as √2/(2√x).

My question is how does √2/(2√x) get simplified to -----> 1/(√2)(√x)
 
\(\displaystyle \dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}}=\dfrac{1}{\sqrt{2}}\)
 
Think of it as "rationalizing the numerator". It is the same idea as "rationalizing the denominator" which is more common.
 
tdotwire8 said:
Hi, I am beginning calculus right now but there's a bit of
Algebra at the end of a limit problem that has me scratching my head. I'm obviously
missing this knowledge as it might be very straight forward, but bear with me here.

So the limit problem is as follows F(x) = √2x No, it is supposed to be √(2x). Use some grouping symbols.


now the answer as I've checked wolframalpha is 1/(√2)(√x) No, it is to be 1/((√2)(√x)), or 1/[(√2)(√x)],
for instance. Again, use grouping symbols.

But really it can be further condensed into 1/√(2x).


I just don't get how this is the answer. Checking the step by step calculation of this derivative
I've follow the exact same until the second last step, this is where the algebra is messing with me.

the last step I get correct is the answer as √2/(2√x).

My [tdotwire8's] question is how does √2/(2√x) get simplified to -----> 1/(√2)(√x)
Click to expand...

You didn't show the steps for how you calculated the derivative. Of different methods,

I can recommend the following one, and it doesn't involve determining that √2/(2√x) = 1/((√2)(√x)).


F(x) = √(2x)


F(x) = \(\displaystyle (2x)^{\frac{1}{2}}\) . . . This is a rewrite.


Then you can use the power rule and the chain rule:


F'(x) = \(\displaystyle \dfrac{1}{2}(2x)^{-\frac{1}{2}}(2)\)


F'(x) = \(\displaystyle (2x)^{-\frac{1}{2}}\)


F'(x) = \(\displaystyle \dfrac{1}{(2x)^{\frac{1}{2}}}\)


F'(x) = \(\displaystyle \dfrac{1}{\sqrt{2x}}\)
 
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tdotwire8 said:
Hi, I am beginning calculus right now but there's a bit of Algebra at the end of a limit problem that has me scratching my head. I'm obviously missing this knowledge as it might be very straight forward, but bear with me here.

So the limit problem is as follows F(x) = √(2x) ....... your problem was probably this or \(\displaystyle F(x) \ = \ \sqrt{2x}\)

now the answer as I've checked wolframalpha is 1/(√2)(√x) ................. if it did, then wolframalpha was wrong. It probably said \(\displaystyle \dfrac{1}{\sqrt{2} \sqrt{x}}\)...... which is correct.

I just don't get how this is the answer. Checking the step by step calculation of this derivative I've follow the exact same until the second last step, this is where the algebra is messing with me.

the last step I get correct is the answer as √2/(2√x).

My question is how does √2/(2√x) get simplified to -----> 1/(√2)(√x)... It does not - not the way you wrote it.
Click to expand...

The way wrote it

1/(√2)(√x) = \(\displaystyle \dfrac{1}{\sqrt{2}} \sqrt{x}\)

I think you wanted to write:

\(\displaystyle \dfrac{1}{\sqrt{2} \sqrt{x}}\)

which should have been written as:

1/[(√2)(√x)] or 1/√(2x)

Carefully look at (check) your grouping symbols to convey correct mathematical expression.
 
Post # 6 is a rehash of what I had already posted a few hours ago in post #5.

edit:


And despite what Wolfram Alpha and www.quickmath.com show, if one allows

radicals/ a radical to be left in the denominator, \(\displaystyle \dfrac{1}{\sqrt{2x}}\)
is the best that can be done as far as simplifying/using fewer symbols.
 
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Thanks for picking out the time to discuss this. It is extremely helpful for me. Thanks for such a valuable help again.
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