Confusion with Calculating Derivative of Inverse Tangent Function

[Saitama]

Homework Statement

This is not a homework question but from a test paper. :)

My exams just got over. In the examination hall, we were made to sit in a particular arrangement. Near me a student of higher class sat near me. They had their maths paper. I am posting one of the question here:-

y=tan^{-1}(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}})
Find \frac{dy}{dx}.

The student near by me was not able to solve this problem so i thought of trying that and helping him. I found the answer by directly applying the formula \frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^2}. ( I don't remember the answer which i got) but when i calculated using wolfram alpha it was completely wrong. I was just embarrassed when i got that answer from wolfram alpha. I kept thinking what the student would be thinking about me. I don't understand why wolfram alpha used u substitution. Why can't we directly apply the formula?

Homework Equations

The Attempt at a Solution

 

[I like Serena]

Hi Pranav-Arora!

I'm afraid you need to apply the chain rule.

This means that if we define:
u(x) = \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}
then:
y=tan^{-1}u
and:
\frac{dy}{dx} = \frac{d}{dx}(tan^{-1}u)=\frac{d}{du}(tan^{-1}u)\frac{du}{dx}

 

[Saitama]

Thanks for the reply ILS!

\frac{dy}{dx} = \frac{d}{dx}(tan^{-1}u)=\frac{d}{du}(tan^{-1}u)\frac{du}{dx}

This is exactly the same thing which i saw at Wolfram Alpha.
But may i know why we have to use the chain rule, why not directly apply the formula? :)

 

[Harrisonized]

The chain rule measures the rate of change of the rate of change.

By the way, the problem is simplified when you multiplied both top and bottom of the fraction by the conjugate of the bottom.

 

[Saitama]

The chain rule measures the rate of change of the rate of change.

By the way, the problem is simplified when you multiplied both top and bottom of the fraction by the conjugate of the bottom.

If i try your method:-
y=tan^{-1}x
x=tan(y)

But when i differentiate with respect to x i i get stuck. What do i get when i differentiate tan(y) with respect to x?

 

[Saitama]

The chain rule measures the rate of change of the rate of change.

By the way, the problem is simplified when you multiplied both top and bottom of the fraction by the conjugate of the bottom.

Ok i did as you said, i get:-
y=tan^{-1}(\frac{1-\sqrt{1-x^4}}{x^2})

So now should i directly apply the formula for differentiating it?

 

[I like Serena]

The chain rule needs to be applied whenever you have a nested expression of x.

Consider for instance f(x) = (x^2+1)^2.
This is equal to f(x) = (x^4+2x^2+1).

If we take the derivative of each we get:
f'(x) = 2(x^2+1) \color{red}{\cdot 2x}
Resp.
f'(x) = 4x^3 + 4x

As you can see the part marked in red is required.

 

[Saitama]

The chain rule needs to be applied whenever you have a nested expression of x.

Consider for instance f(x) = (x^2+1)^2.
This is equal to f(x) = (x^4+2x^2+1).

If we take the derivative of each we get:
f'(x) = 2(x^2+1) \color{red}{\cdot 2x}
Resp.
f'(x) = 4x^3 + 4x

As you can see the part marked in red is required.

Thanks but i knew that. :)

But i am stuck at this:-
y=tan^{-1}(\frac{1-\sqrt{1-x^4}}{x^2})

Should i now directly apply this formula:-
\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^2}

 

[I like Serena]

Uhh, yes?

Yes, you should apply that formula, but you should also apply the chain rule.

 

[Saitama]

It involved too much calculation. I get the answer to be:-
\frac{dy}{dx}=\frac{x}{\sqrt{1-x^4}}

I think its correct. Right?

 

[I like Serena]

It's what wolfram says, but you skipped a few steps.

 

[Saitama]

It's what wolfram says, but you skipped a few steps.

Which steps?

Do you mean that i should post the steps followed by me to solve the problem here?

 

[I like Serena]

It think you want to make a giant shortcut to find the result.

But I'm afraid there's no real shortcut.

You still have to calculate:
\frac {dy} {dx}=\frac {d} {dx} tan^{-1}(\frac{1-\sqrt{1-x^4}}{x^2})<br /> =\frac 1 {1 + (\frac{1-\sqrt{1-x^4}}{x^2})^2} \frac {d} {dx} (\frac{1-\sqrt{1-x^4}}{x^2})

Did you do that?

It's only the sub expressions that you can rewrite a bit to make it less work.
For instance:
\frac {d} {dx} (\frac{1-\sqrt{1-x^4}}{x^2}) =\frac {d} {dx} (\frac{1}{x^2} -\sqrt{\frac 1 {x^4} -1})
makes it a little less work.

And apparently the end result is pretty simple.

 

[Saitama]

You still have to calculate:
\frac {dy} {dx}=\frac {d} {dx} tan^{-1}(\frac{1-\sqrt{1-x^4}}{x^2})<br /> =\frac 1 {1 + (\frac{1-\sqrt{1-x^4}}{x^2})^2} \frac {d} {dx} (\frac{1-\sqrt{1-x^4}}{x^2})

I did exactly the same thing as you said that we have to use chain rule.

It's only the sub expressions that you can rewrite a bit to make it less work.
For instance:
\frac {d} {dx} (\frac{1-\sqrt{1-x^4}}{x^2}) =\frac {d} {dx} (\frac{1}{x^2} -\sqrt{\frac 1 {x^4} -1})
makes it a little less work.

And apparently the end result is pretty simple.

I didn't try that way. Btw, thanks for the tip.

 

[I like Serena]

I did exactly the same thing as you said that we have to use chain rule.

Oh, okay!
I was just wondering what you did to arrive at your result.

 

[HallsofIvy]

Am I the only one that is disturbed by your saying so casually that you were helping a student cheat on his exams?

 

[DivisionByZro]

Am I the only one that is disturbed by your saying so casually that you were helping a student cheat on his exams?

I thought I was the only one startled by that. Sigh... Cheating on an exam and getting a totally wrong answer.