Thought I had it.. nope! Not so easy!

[reardear]

\(\displaystyle y = \sqrt{x^2+1} \:\cdot\: \sqrt[3]{x^3+1} \:\cdot\: \sqrt[4]{x^4+1} \:\cdot\: \:\:...\:\: \cdot \sqrt[100]{x^{100}+1}, \:\:y\prime(1)\:\:\:Evaluate\:exactly.\)
Okay, I thought I would be able to just make a formula but then I saw the \(\displaystyle y\prime(1)\) and "evaluate exactly".

Am I right to assume this is part of following the problem? Do I get the derivative of this and then use summation somehow?
\(\displaystyle {(x^{n}+1})^{\frac{1}{n}}\)

 

[daon2]

Hmmm... What happens if you take the natural log of both sides first?

 

[Deleted member 4993]

\(\displaystyle y = \sqrt{x^2+1} \:\cdot\: \sqrt[3]{x^3+1} \:\cdot\: \sqrt[4]{x^4+1} \:\cdot\: \:\:...\:\: \cdot \sqrt[100]{x^{100}+1}, \:\:y\prime(1)\:\:\:Evaluate\:exactly.\)
Okay, I thought I would be able to just make a formula but then I saw the \(\displaystyle y\prime(1)\) and "evaluate exactly".

Am I right to assume this is part of following the problem? Do I get the derivative of this and then use summation somehow?
\(\displaystyle {(x^{n}+1})^{\frac{1}{n}}\)

And I realize that the \(\displaystyle \sqrt{x^2+1}\) part doesn't fit that formula, so I guess I could figure out the rest then multiply by that.

Why do you think so?

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

 

[reardear]

Hmmm... What happens if you take the natural log of both sides first?

Would I have to do it to the formula?
\(\displaystyle \ln{y} \ = \ln{(x^n \ + \ 1)^{\frac{1}{n}}}\)
\(\displaystyle \ln{y} \ = \frac{1}{n}\ln{(x^n \ + \ 1)}\)

...

\(\displaystyle \frac{1}{y}y\prime \ = \frac{2x}{2(x^2+1)}\)
\(\displaystyle y\prime \ = \frac{2xy}{2(x^2+1)}\)

Why do you think so?

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

Forgot haha

 

[daon2]

\(\displaystyle \ln{y} \ = \frac{1}{n}\ln{(x^n \ + \ 1)}\)

...

\(\displaystyle \frac{1}{y}y\prime \ = \frac{2x}{n(x^n+1)}\)

That 2 is not correct!

I will get you started on a simplified problem.

\(\displaystyle \ln(y) = \ln\left(\sqrt{x^2+1}\sqrt[3]{x^3+1}\sqrt[4]{x^4+1}\right)\)

\(\displaystyle = \frac{1}{2}\ln\left(x^2+1\right)+\frac{1}{3}\ln \left(x^3+1\right)+\frac{1}{4}\ln\left(x^4+1\right)\)

Now take the derivative and simplify. You will notice the right-hand side (when 1 is plugged in) simplifies very nicely, but its 'exact' value will still look unsatisfying to some.

One note: \(\displaystyle y(1) = \sqrt{2}\sqrt[3]{2}\sqrt[4]{2} = 2^{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}\)

 

[reardear]

That 2 is not correct!

I will get you started on a simplified problem.

\(\displaystyle \ln(y) = \ln\left(\sqrt{x^2+1}\sqrt[3]{x^3+1}\sqrt[4]{x^4+1}\right)\)

\(\displaystyle = \frac{1}{2}\ln\left(x^2+1\right)+\frac{1}{3}\ln \left(x^3+1\right)+\frac{1}{4}\ln\left(x^4+1\right)\)

Now take the derivative and simplify. You will notice the right-hand side (when 1 is plugged in) simplifies very nicely, but its 'exact' value will still look unsatisfying to some.

One note: \(\displaystyle y(1) = \sqrt{2}\sqrt[3]{2}\sqrt[4]{2} = 2^{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}\)

Sorry I forgot to change the 2 when I meant to use the formula, but doesn't matter anymore!

\(\displaystyle \Large \frac{x}{x^2+1} + \frac{x^2}{x^3+1} + \frac{x^3}{x^4+1}\)
So plugging in 1, they all equal \(\displaystyle \frac{1}{2}\)!

 

[daon2]

Right, but don't forget that you still need to multiply by y.

 

[reardear]

Right, but don't forget that you still need to multiply by y.

Yeah. So at this point would I use summation for the \(\displaystyle y(1)\) piece you mentioned?

\(\displaystyle \frac{dy}{dx} = y[49.5]\)

 

[daon2]

Yeah. So at this point would I use summation for the \(\displaystyle y(1)\) piece you mentioned?

\(\displaystyle \frac{dy}{dx} = y[49.5]\)

Yup!

 

[reardear]

Yup!

Okay! So I got
\(\displaystyle y = 2^{4.1873775}\)
\(\displaystyle y\prime = 2^{4.1873775}(49.5)\)
\(\displaystyle y\prime = 901.844\)

Not sure how exact I could be since no decimal places were mentioned. Is there a different way I should be doing it?

 

[daon2]

It asked for an exact answer. The only way to state it would be an ugly way, or by means of substitution.

\(\displaystyle \frac{99}{2}\cdot 2^{\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{100}\right)} = 99\cdot 2^{\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{100}\right)-1}\)

The value of \(\displaystyle H_{100}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{100}\) is called the 100th harmonic number. Named so for the harmonic series. So your power of 2 is \(\displaystyle H_{100}-2\). Letting r be this value (and explicitly stating so), you may write the exact answer as \(\displaystyle 99\cdot 2^r\)

 

[reardear]

It asked for an exact answer. The only way to state it would be an ugly way, or by means of substitution.

\(\displaystyle \frac{99}{2}\cdot 2^{\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{100}\right)} = 99\cdot 2^{\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{100}\right)-1}\)

The value of \(\displaystyle H_{100}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{100}\) is called the 100th harmonic number. Named so for the harmonic series. So your power of 2 is \(\displaystyle H_{100}-2\). Letting r be this value (and explicitly stating so), you may write the exact answer as \(\displaystyle 99\cdot 2^r\)

I found out just leaving the fractions in the exponent was good enough ! Thank you for your help!