Question

Differentiate
y=xcos^(-1)x-sqrt(1-x^2).

Answer 1

This looks like an ugly one.

Answer 2

you need to know product rule for the xcos^{-1}(x) part

Answer 3

and chain rule for the sqrt(1-x^2) part

Answer 4

\[\frac{d}{dx}(xcos^{-1}(x))\]
so just looking at this part
do you know how to differentiate this part?

Answer 5

Do we need to use arccos?

Answer 6

cos^(-1)(x) is arccos(x)

Answer 7

\[\frac{d}{dx}\cos^{-1}(x)=\frac{-1}{\sqrt{1-x^2} }\]

Answer 8

@tjb69812 do you know how to apply product rule to the part I mentioned above?

Answer 9

f*g'(x)+f'(x)g

Answer 10

\[\frac{d}{dx} (xcos^{-1}(x))=\cos^{-1}(x) \frac{dx}{dx}+x \frac{dcos^{-1}(x)}{dx}\]

Answer 11

so you should be able to do this part now

Answer 12

\[\frac{d \sqrt{1-x^2}}{dx}=\frac{d(1-x^2)^\frac{1}{2}}{dx} \]

Answer 13

you need to know chain rule

Answer 14

chain rule of the right side?

Answer 15

\[\frac{1}{2}(1-x^2)^{\frac{1}{2}-1}(1-x^2)'\]
look familiar?

Answer 16

yes sir it does

Answer 17

Oooo this is such a cool problem! :D
Simplifies down to a really simple solution!
fun stuff!

Answer 18

guide me along my way!

Answer 19

Oh you didn't figure this one out yet? :U

Answer 20

I'm messing up somewhere it doesn't come out nicely for me :-(

Answer 21

\[\Large\rm y=x \cos^{-1}x-\sqrt{1-x^2}\]
\[\Large\rm \color{royalblue}{y'}=\color{royalblue}{\left(x\right)'} \cos^{-1}x+x\color{royalblue}{\left(\cos^{-1}x\right)'}-\color{royalblue}{\left(\sqrt{1-x^2}\right)'}\]Here is our setup for differentiating, product rule required.
These are the things we need to differentiate, yes?

Answer 22

I would recommend memorizing that the `derivative of sqrt x` is `1 over 2 sqrts of x`. It's one of those functions that shows up so often, that it's worth memorizing instead of applying power rule to.\[\Large\rm \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt x}\]

Answer 23

\[\Large\rm \color{royalblue}{y'}=\color{orangered}{\left(1\right)} \cos^{-1}x+x\color{royalblue}{\left(\cos^{-1}x\right)'}-\color{royalblue}{\left(\sqrt{1-x^2}\right)'}\]You figured out your inverse cosine derivative, yes? :O

Answer 24

can you explain arccos?

Answer 25

\[\Large\rm y=\cos^{-1}x\]\[\Large\rm y'=?\]We need some other approach to figure out this derivative.
We'll rewrite our expression in terms of cosine,\[\Large\rm x=\cos y\]

Answer 26

what rule allows that?

Answer 27

You'll want to brush up on inverse functions maybe.. they're a little tricky if you haven't seen them in a while.

Functions and their inverses have this property:\[\Large\rm f\left[f^{-1}(x)\right]=x\]When you take their composition, you simply get the argument back.
So maybe it would make more sense to apply that to what I was trying to show you. I kind of shortcut'd a bit.\[\Large\rm y=\cos^{-1}(x)\]Apply cosine to both sides,\[\Large\rm \cos (y)=\cos\left[\cos^{-1}(x)\right]\]So our right side simplifies,\[\Large\rm \cos(y)= x\]

Answer 28

yes thank you! i think i can solve from here

Answer 29

Ah ok :)