# Simple differentiation question

• Saitama
In summary: No,not at all.Looking at all the details certainly helps.We've only scratched the surface.It turns out, that it's not the \sqrt{1-\cos^2(\theta)} which is causing the problem.Yes, \sqrt{1-\cos^2(\theta)}=\sqrt{\sin^2(\theta)}=|\sin(\theta)|\,, because if you choose \displaystyle \frac{\pi}{4}<\theta<\frac{3\pi}{4} to correspond to \displaystyle -\frac{1}{\sqrt{}2}<x<\frac{1}{\sqrt{2}}\,, then \sin(\theta)>0\,, so that |\sin
Saitama

## Homework Statement

Find $\frac{dy}{dx}$.
$$y=\sin^{-1}(2x\sqrt{1-x^2}), \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$

## The Attempt at a Solution

I started with substituting x=sinθ.
The expression simplifies to $y=\sin^{-1}(\sin(2θ))$ which is equal to $y=2θ$.
Substituting back the value of θ, i get $y=2\sin^{-1}x$.
Therefore $\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$
According to the book, this answer is correct.
But if a start by substituting x=cosθ. The expression simplifies to $y=2\cos^{-1}θ$, if i differentiate this, i get
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$
I don't understand why i get these two different answers, i suppose it has to do something with the range of x given in the question but i don't seem to get the point.
Any help is appreciated!

Thanks!

Last edited:
Pranav-Arora said:

## Homework Statement

Find $\frac{dy}{dx}$.
$$y=\sin^{-1}(2x\sqrt{1-x^2}), \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$

## The Attempt at a Solution

I started with substituting x=sinθ.
The expression simplifies to $y=\sin^{-1}(\sin(2θ))$ which is equal to $y=2θ$.
Substituting back the value of θ, i get $y=2\sin^{-1}x$.
Therefore $\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$
According to the book, this answer is correct.
But if a start by substituting x=cosθ. The expression simplifies to $y=2\cos^{-1}θ$, if i differentiate this, i get
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$
I don't understand why i get these two different answers, i suppose it has to do something with the range of x given in the question but i don't seem to get the point.
Any help is appreciated!

Thanks!
$\displaystyle \sqrt{\sin^2(\theta)}=|\sin(\theta)|$

I don't understand why you are using a substitution. This problem requires the use of the chain rule and the product rule.

... or you could take the sine of both sides and do implicit differentiation.

SammyS said:
$\displaystyle \sqrt{\sin^2(\theta)}=|\sin(\theta)|$
How does that help? If i take $\displaystyle \sqrt{\sin^2(\theta)}=-\sin(\theta)$, i end up with $\pi+2\cos^{-1}(x)$ and i get the wrong answer as before. :(

Mark44 said:
I don't understand why you are using a substitution. This problem requires the use of the chain rule and the product rule.
Yes, i know about it but don't you see substitution makes it a lot easier. It took some time to write but when i saw the question in my book, i could do it in my mind using substitution.

Pranav-Arora said:
How does that help? If i take $\displaystyle \sqrt{\sin^2(\theta)}=-\sin(\theta)$, i end up with $\pi+2\cos^{-1}(x)$ and i get the wrong answer as before. :(
Well, you didn't show the steps you took in detail, so I thought perhaps this is where you went wrong.

SammyS said:
Well, you didn't show the steps you took in detail, so I thought perhaps this is where you went wrong.

Here are the steps:
Substituting x=cosθ.
$$y=\sin^{-1}(2\cos(\theta)(\sqrt{1-\cos^2(\theta)})$$
$$y=\sin^{-1}(2\cos(\theta)(-\sin(\theta)))$$
$$y=\sin^{-1}(-sin2(\theta))$$
I can rewrite -sin(2θ) as sin(π+2θ).
Therefore, sin-1(sin(π+2θ))=π+2θ=π+2cos-1x.

Please tell me where i am wrong.

Pranav-Arora said:
I can rewrite -sin(2θ) as sin(π+2θ).
Therefore, sin-1(sin(π+2θ))=π+2θ=π+2cos-1x.

Please tell me where i am wrong.
For one thing, $-\sin(2\theta)=\sin(-2\theta)\,,$ which looks like will solve your problem.

Furthermore, to do it your way:

$\sin^{-1}(\sin(\pi+2\theta))\,,$ is some value between $\pi/2\ \text{ and }\ -\pi/2$

In other words $-\pi/2\le\sin^{-1}(\sin(\pi+2\theta))\le\pi/2\,.$

SammyS said:
For one thing, $-\sin(2\theta)=\sin(-2\theta)\,,$ which looks like will solve your problem.
Yes, that solves it. But what about the case when we take $\sqrt{\sin^2(\theta)}=\sin(\theta)$?

Furthermore, to do it your way:

$\sin^{-1}(\sin(\pi+2\theta))\,,$ is some value between $\pi/2\ \text{ and }\ -\pi/2$

In other words $-\pi/2\le\sin^{-1}(\sin(\pi+2\theta))\le\pi/2\,.$
How this relation helps out? I know about this but i am still at a loss in understanding how this could help me out.

Pranav-Arora said:
Yes, that solves it. But what about the case when we take $\sqrt{\sin^2(\theta)}=\sin(\theta)$?
Well, you have this same problem with your first substitution: $x=\sin(\theta)\,.$ It's just that in this case the positive square root works out correctly.

$2x\sqrt{1-x^2}$ becomes

$2\sin(\theta)\sqrt{1-\sin^2(\theta)}$
$=2\sin(\theta)\sqrt{\cos^2(\theta)}$

$=2\sin(\theta)(\pm\cos(\theta))$

$=\pm\sin(2\theta)$​
Often with trigonometry expressions, a '±' symbol indicates that you need to choose the correct sign based on context on the situation, rather than in solving algebraic equations where the '±' symbol means that each case is valid.

SammyS said:
Often with trigonometry expressions, a '±' symbol indicates that you need to choose the correct sign based on context on the situation, rather than in solving algebraic equations where the '±' symbol means that each case is valid.

So, how would i know which sign to choose here? I have a couple of more problems relating to the same problem where i get two answers. Sorry, if i am acting like a dumb. :uhh:

Pranav-Arora said:
So, how would i know which sign to choose here? I have a couple of more problems relating to the same problem where i get two answers. Sorry, if i am acting like a dumb. :uhh:
No,not at all.

Looking at all the details certainly helps.

We've only scratched the surface.

It turns out, that it's not the $\sqrt{1-\cos^2(\theta)}$ which is causing the problem.

Yes, $\sqrt{1-\cos^2(\theta)}=\sqrt{\sin^2(\theta)}=|\sin(\theta)|\,,$ because if you choose $\displaystyle \frac{\pi}{4}<\theta<\frac{3\pi}{4}$ to correspond to $\displaystyle -\frac{1}{\sqrt{}2}<x<\frac{1}{\sqrt{2}}\,,$ then $\sin(\theta)>0\,,$ so that $|\sin(\theta)|=\sin(\theta)\ .$

The problem comes from treating $\sin^{-1}(\sin(2\cos^{-1}(x)))$ too casually. After all, $\theta=\cos^{-1}(x)\ .$

But it's late here & I couldn't sleep, so I looked further into this problem, but I really need to try to sleep now.

I'll get back to this later.

Cheers.

Last edited:
SammyS said:
No,not at all.

Looking at all the details certainly helps.

We've only scratched the surface.

It turns out, that it's not the $\sqrt{1-\cos^2(\theta)}$ which is causing the problem.

Yes, $\sqrt{1-\cos^2(\theta)}=\sqrt{\sin^2(\theta)}=|\sin(\theta)|\,,$ because if you choose $\displaystyle \frac{\pi}{4}<\theta<\frac{3\pi}{4}$ to correspond to $\displaystyle -\frac{1}{\sqrt{}2}<x<\frac{1}{\sqrt{2}}\,,$ then $\sin(theta)>0\,,$ so that $|\sin(\theta)|=\sin(\theta)\ .$
Ok, i understand till here. Thanks for the explanation.

The problem comes from treating $\sin^{-1}(\sin(2\cos^{-1}(x)))$ too casually. After all, $\theta=\cos^{-1}(x)\ .$

But it's late here & I couldn't sleep, so I looked further into this problem, but I really need to try to sleep now.

I'll get back to this later.

Cheers.
I can wait for this. Its afternoon here so i have a whole day to prepare for my mathematics exam tomorrow. I am done with all the problems except this one. I guess you will be awake by night(here). Thanks for all the help! I will see if i could myself solve this.

Good night SammyS.

I may be doing just small bits at a time.

While it's always true that $\sin(\sin^{-1}(x))=x\ ,,$ as long as $\sin^{-1}(x)$ is defined, One has to be careful when going the other way. $\sin^{-1}(\sin(\theta))=\theta$ only for $\displaystyle -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\ .$

It looks like you were careful about that in regards to the ends of your interval.

but ...

SammyS said:
but ...

I have got it finally. Here we have $\sin^{-1}(\sin(2\theta))$ where $2\theta$ lies between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. I visualized what you said using a graph and i was stunned by seeing how a big fool i am. For this range, the value of function $\sin^{-1}(\sin(x))$ is equal to $\pi-2x$.
I have got it now, thank you very much. You saved me.

Pranav-Arora said:
I have got it finally. Here we have $\sin^{-1}(\sin(2\theta))$ where $2\theta$ lies between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. I visualized what you said using a graph and i was stunned by seeing how a big fool i am. For this range, the value of function $\sin^{-1}(\sin(x))$ is equal to $\pi-2x$.
I have got it now, thank you very much. You saved me.
Your no fool! It took me a while to see where the problem was arising.

Glad to help! It's good to work with someone willing to think outside the box, and willing to explore a bit on his/her own!
Those "inverse" trig functions can be tricky, what with their limited range & all.

I too used some graphs to get my head around this problem!

Did you happen look at the graph of y if you extend the domain of the problem to -1 ≤ x ≤ 1 ?

SammyS said:
Did you happen look at the graph of y if you extend the domain of the problem to -1 ≤ x ≤ 1 ?
Why? I just looked at it on wolframalpha.

Pranav-Arora said:
Why? I just looked at it on wolframalpha.
Well, there's those sharp cusps at x = ±1/√2 , so, a big discontinuity in the first derivative there.

--- just thought that was interesting.

SammyS said:
Well, there's those sharp cusps at x = ±1/√2 , so, a big discontinuity in the first derivative there.

--- just thought that was interesting.

Yeah, just saw those.

Anyways, today was my exam and any question related to this wasn't asked. But i lost marks in matrices.

Pranav-Arora said:
Yeah, just saw those.

Anyways, today was my exam and any question related to this wasn't asked. But i lost marks in matrices.

SammyS said:

Yes, i hope so.

## What is simple differentiation?

Simple differentiation is a mathematical process that involves finding the rate of change of a function with respect to its independent variable. It is a fundamental concept in calculus and is used to determine the slope of a curve at a particular point.

## What is the formula for simple differentiation?

The formula for simple differentiation is dy/dx, which represents the derivative of a function y with respect to its independent variable x. This can also be written as f'(x) or df/dx.

## What are the steps for simple differentiation?

The steps for simple differentiation are as follows:

1. Identify the function to be differentiated.
2. Rewrite the function using the power rule, product rule, quotient rule, or chain rule if necessary.
3. Differentiate each term of the function using the appropriate rule.
4. Combine like terms and simplify the resulting expression.

## What are the applications of simple differentiation?

Simple differentiation has many applications in science, engineering, and economics. It is used to analyze and model rates of change, such as velocity, acceleration, and growth. It is also used in optimization problems to find the maximum or minimum value of a function.

## What are some common mistakes when performing simple differentiation?

Some common mistakes when performing simple differentiation include forgetting to apply the chain rule, making arithmetic errors, and incorrectly using the power, product, or quotient rule. It is important to carefully follow the steps and double check your work to avoid these mistakes.

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