Question

derivative of:

postin. postin pic. My question is though, why on the third line, is it [1+(1-x^2)-x^2]? like where did the -x^2 come from?

Answer 1

they found common denominators which is \[\sqrt{1-x}\] and combined the two terms. This allows it to be reduced more.

Answer 2

\(\large y = arcsinx + x\sqrt{1-x^2}\)

\(\large y' = \frac{1}{\sqrt{1-x^2}} + \sqrt{1-x^2} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

Answer 3

fine till here right ?

Answer 4

I'm sorry combined which two terms? and @ganeshie8 I got up to there it's the next line I'm confused about. i sorta understand the common denim. but I'm not too certain

Answer 5

\(\large y = arcsinx + x\sqrt{1-x^2}\)

\(\large y' = \frac{1}{\sqrt{1-x^2}} + \sqrt{1-x^2} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

\(\large y' = (\sqrt{1-x^2})^{\frac{-1}{2}} + \sqrt{1-x^2} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

Answer 6

still fine ?

Answer 7

yes

Answer 8

ive made a mistake there,

it shoudl be :-

\(\large y = arcsinx + x\sqrt{1-x^2}\)

\(\large y' = \frac{1}{\sqrt{1-x^2}} + \sqrt{1-x^2} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

\(\large y' = (1-x^2)^{\frac{-1}{2}} + \sqrt{1-x^2} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

Answer 9

next factor \((1-x^2)^\frac{-1}{2}\) from each term

Answer 10

oh ok i see the square root.

Answer 11

yes :)

Answer 12

from the entire equ?

Answer 13

\(\large y = arcsinx + x\sqrt{1-x^2}\)

\(\large y' = \frac{1}{\sqrt{1-x^2}} + \sqrt{1-x^2} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

\(\large y' = (1-x^2)^{\frac{-1}{2}} + \sqrt{1-x^2} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

\(\large y' = (1-x^2)^{\frac{-1}{2}} + (1-x^2)^{1 - \frac{1}{2}} + x(\frac{1}{2}) (1-x^2)^{\frac{-1}{2}}(-2x)\)

Answer 14

yes from the entire right side, factor that..

Answer 15

ok I get it. Thank you!!

Answer 16

Thanks for the help guys(: appreciate it

Answer 17

np :)