I need help!

K

Katener88

New member
Joined
Mar 13, 2012
Messages
2
I am teaching myself Calc, and I'm stuck on this Chain Rule problem!
Sqrt((x^2+1)(sqrtx+1)^3) help someone!
I have the entire bottom part correct, but the top is really confusing. All the derivation is complete, but I can't get it to properly simplify
 
Last edited:
Your problem is \(\displaystyle \sqrt{\frac{x^{2}+1}{(\sqrt{x}+1)^{3}}}\)

Or equivalently \(\displaystyle \frac{\sqrt{x^{2}+1}}{\sqrt{(\sqrt{x}+1)^{3}}}\)

\(\displaystyle =\frac{(x^{2}+1)^{\frac{1}{2}}}{(\sqrt{x}+1)^{\frac{3}{2}}}\)

We can write this as \(\displaystyle \sqrt{x^{2}+1}(\sqrt{x}+1)^{\frac{-3}{2}}\)

Now, instead of the quotient rule, use the product rule.

\(\displaystyle (fg)'=fg'+gf'\)

Let \(\displaystyle f=(x^{2}+1)^{\frac{1}{2}}, \;\ g=(\sqrt{x}+1)^{\frac{-3}{2}}\)

\(\displaystyle f'=\frac{1}{2}2x(x^{2}+1)^{\frac{-1}{2}}, \;\ g'=\frac{-3}{2}(\sqrt{x}+1)^{\frac{-5}{2}}\cdot \frac{1}{2\sqrt{x}}\)

With problems like this, the chain rule is essentially the derivative of the inside times the derivative of the outside.

Take the derivative of \(\displaystyle (x^{2}+1)^{\frac{1}{2}}\) for instance,

The derivative of the outside is \(\displaystyle \frac{1}{2}(x^{2}+1)^{\frac{-1}{2}}\)

The derivative of the inside is the derivative of \(\displaystyle x^{2}+1\), which is \(\displaystyle 2x\).

Now, take the derivative of \(\displaystyle (\sqrt{x}+1)^{\frac{-3}{2}}\)

The derivative of the outside is \(\displaystyle \frac{-3}{2}(\sqrt{x}+1)^{\frac{-5}{2}}\)

The derivative of the inside is the derivative of \(\displaystyle \sqrt{x}+1\), which is \(\displaystyle \frac{1}{2\sqrt{x}}\).

So, we get upon multiplying them:

\(\displaystyle \frac{-3}{2}(\sqrt{x}+1)^{\frac{-5}{2}}\cdot \frac{1}{2}x^{\frac{-1}{2}}\)

Which simplifies to \(\displaystyle \frac{-3}{4}x^{\frac{-1}{2}}(\sqrt{x}+1)^{\frac{-5}{2}}\)

Or you can write it as \(\displaystyle \frac{-3}{4\sqrt{x}(\sqrt{x}+1)^{\frac{5}{2}}}\)

Now, can you put it all together?. It's rather messy.

\(\displaystyle (x^{2}+1)^{\frac{1}{2}}\cdot \frac{-3}{4}x^{\frac{-1}{2}}(\sqrt{x}+1)^{\frac{-5}{2}}+(\sqrt{x}+1)^{\frac{-3}{2}}\cdot \frac{1}{2}2x(x^{2}+1)^{\frac{-1}{2}}\)

Now, simplify this down to some form. Books often times have things simplified to an unrecognizable form.

The algebra is usually the worst part of calculus for students.

But, we can hammer into some shape:

\(\displaystyle \frac{x^{2}+4x^{\frac{3}{2}}-3}{4(\sqrt{x}+1)^{\frac{5}{2}}\sqrt{x(x^{2}+1)}}\)
 
Last edited:
galactus said:
Your problem is \(\displaystyle \sqrt{\frac{x^{2}+1}{(\sqrt{x}+1)^{3}}}\)

Or equivalently \(\displaystyle \frac{\sqrt{x^{2}+1}}{\sqrt{(\sqrt{x}+1)^{3}}}\)
Click to expand...

I don't understand why the division sign is there? It doesn't say it in the original question?
 
If there is no division sign (may have been corrected by OP after Galactus started the solution):

\(\displaystyle f(x) \ = \ \sqrt{(x^2+1) \ * \ (\sqrt{x} \ + \ 1)^3}\)

\(\displaystyle f(x) \ = \ (\sqrt{x} \ + \ 1) \ * \ \sqrt{(x^2+1) \ * \ (\sqrt{x} \ + \ 1)}\)

\(\displaystyle f(x) \ = \ (\sqrt{x} \ + \ 1) \ * \ \sqrt{x^{\frac{5}{2}} \ + \ x^2 \ + \ \sqrt{x} \ + \ 1} \ = \ f_1(x) \ * \ f_2(x) \)

\(\displaystyle f'_1(x) \ = \ \dfrac{1}{2\sqrt{x}} \)

\(\displaystyle f'_2(x) \ = \ \dfrac{1}{2\left [\sqrt{x^{\frac{5}{2}} \ + \ x^2 \ + \ \sqrt{x} \ + \ 1}\right ]}\left [\frac{5}{2} \ * \ x^{\frac{3}{2}} \ + 2*\ x \ + \ \frac{1}{2\sqrt{x}}\right ]\)

\(\displaystyle f'(x) \ = \ f'_1(x) \ * \ f_2(x) \ + \ f_1(x) \ * \ f'_2(x)\)

Now continue....
 
It would help if katener88 would come back and tell us what was intended!
There was a remark about the "denominator" but no fraction in the given function.
 
I suppose the OP is referring to the denominator/s of the derivative.
 
Uhmm..

galactus said:
Your problem is \(\displaystyle \sqrt{\frac{x^{2}+1}{(\sqrt{x}+1)^{3}}}\)

Or equivalently \(\displaystyle \frac{\sqrt{x^{2}+1}}{\sqrt{(\sqrt{x}+1)^{3}}}\)

\(\displaystyle =\frac{(x^{2}+1)^{\frac{1}{2}}}{(\sqrt{x}+1)^{\frac{3}{2}}}\)

We can write this as \(\displaystyle \sqrt{x^{2}+1}(\sqrt{x}+1)^{\frac{-3}{2}}\)

Now, instead of the quotient rule, use the product rule.

\(\displaystyle (fg)'=fg'+gf'\)

Let \(\displaystyle f=(x^{2}+1)^{\frac{1}{2}}, \;\ g=(\sqrt{x}+1)^{\frac{-3}{2}}\)

\(\displaystyle f'=\frac{1}{2}2x(x^{2}+1)^{\frac{-1}{2}}, \;\ g'=\frac{-3}{2}(\sqrt{x}+1)^{\frac{-5}{2}}\cdot \frac{1}{2\sqrt{x}}\)

With problems like this, the chain rule is essentially the derivative of the inside times the derivative of the outside.

Take the derivative of \(\displaystyle (x^{2}+1)^{\frac{1}{2}}\) for instance,

The derivative of the outside is \(\displaystyle \frac{1}{2}(x^{2}+1)^{\frac{-1}{2}}\)

The derivative of the inside is the derivative of \(\displaystyle x^{2}+1\), which is \(\displaystyle 2x\).

Now, take the derivative of \(\displaystyle (\sqrt{x}+1)^{\frac{-3}{2}}\)

The derivative of the outside is \(\displaystyle \frac{-3}{2}(\sqrt{x}+1)^{\frac{-5}{2}}\)

The derivative of the inside is the derivative of \(\displaystyle \sqrt{x}+1\), which is \(\displaystyle \frac{1}{2\sqrt{x}}\).

So, we get upon multiplying them:

\(\displaystyle \frac{-3}{2}(\sqrt{x}+1)^{\frac{-5}{2}}\cdot \frac{1}{2}x^{\frac{-1}{2}}\)

Which simplifies to \(\displaystyle \frac{-3}{4}x^{\frac{-1}{2}}(\sqrt{x}+1)^{\frac{-5}{2}}\)

Or you can write it as \(\displaystyle \frac{-3}{4\sqrt{x}(\sqrt{x}+1)^{\frac{5}{2}}}\)

Now, can you put it all together?. It's rather messy.

\(\displaystyle (x^{2}+1)^{\frac{1}{2}}\cdot \frac{-3}{4}x^{\frac{-1}{2}}(\sqrt{x}+1)^{\frac{-5}{2}}+(\sqrt{x}+1)^{\frac{-3}{2}}\cdot \frac{1}{2}2x(x^{2}+1)^{\frac{-1}{2}}\)

Now, simplify this down to some form. Books often times have things simplified to an unrecognizable form.

The algebra is usually the worst part of calculus for students.

But, we can hammer into some shape:

\(\displaystyle \frac{x^{2}+4x^{\frac{3}{2}}-3}{4(\sqrt{x}+1)^{\frac{5}{2}}\sqrt{x(x^{2}+1)}}\)
Click to expand...




I'm not sure how you got that at all!?!
 
You must log in or register to reply here.
Top