I just get lost after applying the derivatives in the expressionKeep going. You are on the right path.
Now just use those y' and y" and evaluate the given expression. It is a dreary calculation but doable.We have what y is (first line). We are supposed to calculate the expression under it (second line). It gets too complicated. Is there a trick? View attachment 32257
Let [imath]f(x)=x\quad \&\quad g(x)=\sqrt{x^2+1}[/imath]We have what y is (first line). We are supposed to calculate the expression under it (second line). It gets too complicated. Is there a trick? View attachment 32257
No, I have 4 possible answersNow just use those y' and y" and evaluate the given expression. It is a dreary calculation but doable.
Do you have the expected answer?
I did it again. I got 0. Is it correct? View attachment 32262
It's just simplify. I don't see the answer.simplify (1+x)^2(n*(sqrt(x^2+1)+x)^n*(n*sqrt(x^2+1)-x))/(x^2+1)^(3/2)+x(n*(sqrt(x^2+1)+x)^n)/sqrt(x^2+1)-n^2(1+sqrt(x^2+1))^n - Wolfram|Alpha
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Oh I forgot *0 in the last lineNo, the last line is horribly wrong.
Basically you are saying that if 9*0=0, then 9 = 0. NO! 9=9, 9 does not equal 0!
OK!Oh I forgot *0 in the last line
so it is 0?