# Question about Inverse Derivative Hyperbola function

• I
• Vividly
In summary, the conversation is about solving an equation and the use of the hyperbola formula. The person is confused and made a mistake in their calculations, particularly with the square root part. They realize they should have squared the y^2 value at the end to eliminate the square root. The other person advises them to learn the proper way of doing calculus.

#### Vividly

TL;DR Summary
Im confused about a certain part of solving an equation. So I used the hyerbola formula to find the answer but I think I did the math wrong.
X^2-y^2=c^2
X=1
Y= (2x^5-1)^2
I did the calculations as you can see in the picture but I know I messed up on the square root part. When you square one side you have to square the whole other side.
So Far I got
1^2-((2x^5-1)^2)^2 =c^2
And as you can see in the picture the math is incorrect due to the process that's suppose to be done in math. When I square both sides I get a different answer when I simplify.
As you can see on the paper that where I am stuck because the answer is 5/2x-2x^6

#### Attachments

Vividly said:

Im confused about a certain part of solving an equation. So I used the hyerbola formula to find the answer but I think I did the math wrong.
X^2-y^2=c^2
X=1
Y= (2x^5-1)^2
I did the calculations as you can see in the picture but I know I messed up on the square root part. When you square one side you have to square the whole other side.
So Far I got
1^2-((2x^5-1)^2)^2 =c^2
And as you can see in the picture the math is incorrect due to the process that's suppose to be done in math. When I square both sides I get a different answer when I simplify.
As you can see on the paper that where I am stuck because the answer is 5/2x-2x^6
As I mentioned on your previous post, it's not possible to keep trying to do calculus this way. Triangles and graphs get you started, but eventually you should learn the chain rule and the algebra of differentiation.

When you go wrong, then you are asking a lot for someone to take the time and effort to analyse this personal and peculiar approach - since no one else does it this way.

Perhaps someone else will have the time to look at this, but my advice is to start doing calculus properly.

There are different ways of doing math. As long as you follow the logic you can solve the problem. I do math in a way that I understand. I figured out what I did wrong. I was squaring the y^2 value to early in the problem. I was suppose to do it at the end to eliminate the square root.