# Math Help - A Few questions on differentiation

1. ## A Few questions on differentiation

Hi
I need help on the following:

1)Differentiate with respect with x: $sin^2(x)$

I got $2sin(x)cos(x)$, however i don't understand why the book's answer is $sin(2x)$.

2)Differentiate with respect with x: $arcos(\sqrt{1-x^2})$

This is what i have done:

$\frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}$

$= \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}$

What would i do next?

3) Prove that $\frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}$

How would i approach this??

P.S

2. Originally Posted by Paymemoney
Hi
I need help on the following:

1)Differentiate with respect with x: $sin^2(x)$

I got $2sin(x)cos(x)$, however i don't understand why the book's answer is $sin(2x)$.

2)Differentiate with respect with x: $arcos(\sqrt{1-x^2})$

This is what i have done:

$\frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}$

$= \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}$

What would i do next?

3) Prove that $\frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}$

How would i approach this??

P.S
$sin(2x) = 2 sinx. cosx$ So your answer is correct!

3. ## clarification

Originally Posted by Paymemoney
Hi
I need help on the following:

1)Differentiate with respect with x: $sin^2(x)$

I got $2sin(x)cos(x)$, however i don't understand why the book's answer is $sin(2x)$.

2)Differentiate with respect with x: $arcos(\sqrt{1-x^2})$

This is what i have done:

$\frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}$

$= \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}$

What would i do next?

3) Prove that $\frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}$

How would i approach this??

P.S
1.sin(2x)=sin(x+x)=sinxcosx+cosxsinx=2sinxcosx

4. ## a better way

Originally Posted by Paymemoney
Hi
I need help on the following:

1)Differentiate with respect with x: $sin^2(x)$

I got $2sin(x)cos(x)$, however i don't understand why the book's answer is $sin(2x)$.

2)Differentiate with respect with x: $arcos(\sqrt{1-x^2})$

This is what i have done:

$\frac{dy}{dx} = \frac{-0.5(1-x^2)^{-0.5} * 2x}{\sqrt{1-(\sqrt{1-x^2})^2}}$

$= \frac{x}{\sqrt{1-x^2}\sqrt{x^2}}$

What would i do next?

3) Prove that $\frac{d}{dx}(arccos(x) = \frac{-1}{\sqrt{1-x^2}}$

How would i approach this??

P.S
2.there is a better way to do the second question:let y=arc(cos(sqrt(1-x^2))).then cosy=sqrt(1-x^2)=>-sinydy/dx=((-1/2)2x)/sqrt(1-x^2).find out siny from the original expression. plug in the value.things should get cancelled out and you should get your answer.

5. 3. Just in case a picture helps...

... where

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

_________________________________________
Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

6. i still don't understand how to do question 3. how would i start it?

7. OK, in this version, you start with y = arccos x, top left.

Substitute (or map to) cos theta for x, then differentiate with respect to theta using the chain rule. And this is equal to differentiating theta with respect to theta. So...

$\frac{dy}{d\theta}\ =\ \frac{dy}{dx} \frac{dx}{d\theta}\ =\ 1$

$\Rightarrow\ \frac{dy}{dx}\ =\ \frac{1}{\frac{dx}{d\theta}}\ =\ \frac{-1}{\sin \theta}$

So, using Pythag,

$\Rightarrow\ \frac{dy}{dx}\ =\ \frac{-1}{\sqrt{1 - \cos^2 \theta}}$

Then back-substitute x for cos theta.

$y =$ arccos $x\ \Rightarrow\ x = \cos y$

Then differentiate with respect to x...

Substitute arccos x for y...

$\Rightarrow\ \frac{dy}{dx}\ =\ \frac{-1}{\sin (\arccos x)}$

Then use a right-triangle diagram or look up at http://staff.jccc.net/swilson/trig/compositions.htm to simplify.

8. so basically i make $x=cos\theta$
which then i substitute into original equation:

$= arccos(cos\theta)$

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

sub back into equation and i get this:
$=\frac{-1}{\sqrt{1-x^2}}$

9. Originally Posted by Paymemoney
so basically i make $x=cos\theta$
which then i substitute into original equation:

$arccos(cos\theta)$

$\Rightarrow \frac{d}{dx} \arccos x = = \frac{-1}{\sqrt{1-cos^2(\theta)}}" alt="\Rightarrow \frac{d}{dx} \arccos x = = \frac{-1}{\sqrt{1-cos^2(\theta)}}" />

sub back into equation and i get this:
$=\frac{-1}{\sqrt{1-x^2}}$
With slight correction, yes

10. Originally Posted by Pulock2009
2.there is a better way to do the second question:let y=arc(cos(sqrt(1-x^2))).then cosy=sqrt(1-x^2)=>-sinydy/dx=((-1/2)2x)/sqrt(1-x^2).find out siny from the original expression. plug in the value.things should get cancelled out and you should get your answer.
can you explain how you got this, i don't understand.

11. Originally Posted by Paymemoney
can you explain how you got this, i don't understand.
I think (s)he is referring to what I described in my second post as 'a different way...'

Edit:

Maybe not, though...