/sci/ - Science & Math

[Anonymous No.3164063]

f(x) = sqrt(1+sqrt(x))
f'(x) = ?

The answer wolfram alpha gives me differs from the answer the book gives me.

Book solution: f'(x) = 1/(4sqrt(x+sqrt(x)))
Wolfram alpha solution: f'(x) = 1/(4sqrt(sqrt(x)+1)sqrt(x))

Can anybody please show me the steps? I am able to arrive at the solution wolfram alpha gives me, so the steps from there to the final solution would be great.

I suck so fucking hard at LaTeX, but I'll give it a (probably failed) try here:

Book solution:

<span class="math">\frac{1}{4\sqrt{x+\sqrt{x}}}[/spoiler]

Wolfram alpha solution:

<span class="math">\frac{1}{4\sqrt{\sqrt{x}+1}\sqrt{x}}[/spoiler]

Picture unrelated.

[Anonymous No.3164079]

>>3164063
What do you know, it actually worked for once.

Here's the original equation in LaTeX:

<span class="math">f(x)=\sqrt{1+\sqrt{x}}[/spoiler]

[Anonymous No.3164073]

learn to multiply sqrts

sqrt(a)sqrt(b) = sqrt(ab)

now apply that to your second solution

[Anonymous No.3164086]

>The answer wolfram alpha gives me differs from the answer the book gives me.
only as much as x+x^2 and x(1+x) are different.

[mtp !HjLTtX.XQ2 No.3164100]

>>3164063
f(x) = sqrt(1+sqrt(x))differentiate the outside differentiate the inside
f'(x)=1/2 * 1/2 *(1+sqrt(x))^(-0.5)= 1/(4sqrt(1+sqrt(x))) wchich is the books solution
i dont know where WA solution comes from, but then i dont know a lot about maths.

[mtp !HjLTtX.XQ2 No.3164111]

>>3164100
ah fuck, it just hit me.
when you differentiate the inside you get 0.5x^-0.5, which is consistent with wolframs answer, the book is wrond they just forgot how to differentiate x^0.5
according to them d/dx(x^0.5) = 0.5, but it equals 0.5x^-0.5

[Anonymous No.3164116]

>>3164111
>>3164100
>i dont know a lot about maths.

got it in 1

the two answers are as different as 3*2 is different to 6

[Anonymous No.3164124]

>>3164111
lol
my book got the answer 1+2,
but wolfram alpha got the answer 2+1.
wat do?

[Anonymous No.3164129]

Book is wrong, WolframAlpha is right. Don't listen to the people that tell you they're the same. They're idiots that don't know how to distribute properly.

[Anonymous No.3164135]

>>3164129
1/10

[Anonymous No.3164143]

>>3164135
No, seriously. When you distribute the <span class="math">\sqrt{x}[/spoiler] into the radical, it doesn't work the same as distributing <span class="math">x(x+1)[/spoiler]. The radical forces you to square the <span class="math">\sqrt{x}[/spoiler] inside the radical when you distribute.

[Anonymous No.3164166]

>>3164143
Cont'd.

You see, the answers in the backs of math textbooks are written by graduate students. These graduate students like to take shortcuts, and they sometimes make mistakes. As proof that I'm not crazy or trolling, I offer this:

http://www.wolframalpha.com/input/?i=1%2F%284*sqrt%28x%2Bsqrt%28x%29%29%29+%3D+1%2F%284*sqrt%28%28sq
rt%28x%29%2B1%29%29sqrt%28x%29%29

[Anonymous No.3164175]

>>3164166
you're right!
Well I'll be.

I trust wolfram more than textbooks, so I assume wolfram is correct.

[mtp !HjLTtX.XQ2 No.3164177]

ITT: trolls trolling trolled trolls. oh and OP.

>pic related: this thread

[Anonymous No.3164180]

>>3164063
>book solution: f'(x) = 1/(4sqrt(x+sqrt(x)))

should read 1/(4sqrt(x+x sqrt(x)))

[Anonymous No.3164190]

>>3164073
OP here.

<span class="math">\sqrt{\sqrt{x}+1}\sqrt{x}=(x^\frac{1}{2}+1)^\frac{1}{2}(x)^\frac{1}{2}=({x}^{\frac{1}{2}\cdot\frac{1
}{2}}+1^\frac{1}{2})(x)^\frac{1}{2}=(x^\frac{1}{4}+1^\frac{1}{2})(x^\frac{1}{2})={x}^{\frac{1}{4}+\f
rac{1}{2}}+x^\frac{1}{2}=x^\frac{3}{4}+x^\frac{1}{2}...[/spoiler]

This is about all I can do, it seems to my math-retarded mind.

[Anonymous No.3164198]

>>3164073

<span class="math">\sqrt{\sqrt{x}+1}\sqrt{x}=(x^\frac{1}{2}+1)^\frac{1}{2}(x)^\frac{1}{2}=({x}^{\frac{1}{2}\cdot\frac{1
}{2}}+1^\frac{1}{2})(x)^\frac{1}{2}=(x^\frac{1}{4}+1^\frac{1}{2})(x^\frac{1}{2})={x}^{\frac{1}{4}+\f
rac{1}{2}}+x^\frac{1}{2}=x^\frac{3}{4}+x^\frac{1}{2}...[/spoiler]

What to do here?

[Anonymous No.3164202]

>>3164198
God fucking damnit, why do I keep getting this error? Gonna try once more:

<span class="math">\sqrt{\sqrt{x}+1}\sqrt{x}=(x^\frac{1}{2}+1)^\frac{1}{2}(x)^\frac{1}{2}=({x}^{\frac{1}{2}\cdot\frac{1
}{2}}+1^\frac{1}{2})(x)^\frac{1}{2}=(x^\frac{1}{4}+1^\frac{1}{2})(x^\frac{1}{2})={x}^{\frac{1}{4}+\f
rac{1}{2}}+x^\frac{1}{2}=x^\frac{3}{4}+x^\frac{1}{2}[/spoiler]

Plain text:

\sqrt{\sqrt{x}+1}\sqrt{x}=(x^\frac{1}{2}+1)^\frac{1}{2}(x)^\frac{1}{2}=({x}^{\frac{1}{2}\cdot\frac{1
}{2}}+1^\frac{1}{2})(x)^\frac{1}{2}=(x^\frac{1}{4}+1^\frac{1}{2})(x^\frac{1}{2})={x}^{\frac{1}{4}+\f
rac{1}{2}}+x^\frac{1}{2}=x^\frac{3}{4}+x^\frac{1}{2}

[Anonymous No.3164221]

>>3164202
Can somebody help me to get the LaTeX displayed so I can get some help with my algebra, pretty please?

[Anonymous No.3164295]

>>3164221
Or maybe someone can show me how Wolfram Alpha got the alternate form?

http://www.wolframalpha.com/input/?i=x^%281%2F2%29%2Bx^%283%2F4%29