Question

Find the derivative of the function:
y = arctan(2x^2-3)

Answer 1

I have most of the problem done already, it's just one part that's messing me up. The algebra ironically, lol.

Answer 2

solve by first principles, or by takinga tangent of both sides?

Answer 3

I took the tangent of both sides.

Answer 4

\[(\arctan(u))'=\frac{ u' }{ 1+u ^{2} }\]

Answer 5

Take u=2x²-3

Answer 6

tan y =f(x)

sec^2(y) y' = f'

y' = f'/(tan^2(y) + 1)

Answer 7

but tan(y) = f(x) sooo

y' = f(x)/([f(x)]^2+1)

Answer 8

lost a ' on top lol

Answer 9

\[u = 2x-3\]\[u' = 2\]

Answer 10

so that hard part you say is:

4x
-------------
(2x^2-3)^2 + 1

Answer 11

Yeah, and replace on the formula.

Answer 12

It's 2x^2-3 btw.

Answer 13

Yeahm like I went and plug that in and got: \[y' = \frac{ 2 }{ 1+(2x^2-3)^2 }\]

Answer 14

Unless I did something wrong there.

Answer 15

Should be 4x on top.

Answer 16

... I left off the the exponent of u... :/

Answer 17

Let me to re-do that, actually... one sec. And it'll help mwe with learning this.

Answer 18

Ok take your time.

Answer 19

@math&ing001 -- But, yes, that is where I'm stuck. \[y' =\frac{ 4x }{ 1+(2x^2-3)^2 }\]

Answer 20

You need to simplify it ?

Answer 21

Yeah.

Answer 22

I'd start with the term (2x^2-3)^2

Answer 23

That would just be using the FOIL method? My Algebra skills are just rusty, not to mention Calculus. But, I'm willing to work through this, haha

Answer 24

You could use FOIL or the formula for binomials : (a+b)^2 = a^2 + 2ab + b^2

Answer 25

Ahhh, okay. I had used FOIL and got: \[(2x^2-3)^2 = 4x^4-12x^2+9\]

Answer 26

Yep that's correct !

Answer 27

@MoonlitFate

Answer 28

So that would give us: \[y'=\frac{ 4x }{ 4x ^{4}-12x ^{2}+10 }\] I don't think it can be further simplified.

Answer 29

Oh, really? Maybe that's why... I'm trying to figure out an answer out, and keep coming up with nothing.