Question

differentiate

Answer 1

\[(1+sinx)/(x+cosx)\]

Answer 2

@AllTehMaffs

Answer 3

Do you know how to differentiate 1+sinx?

Answer 4

wouldnt it just be cosx

Answer 5

Yes.
What about (x+cosx)?

Answer 6

x-sinx

Answer 7

No...
\[\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)\]

Answer 8

nvr mind it would be 1-sinx right?

Answer 9

Yes, for x+cosx.

Now, let u(x) = (1+sinx), v(x) = (x+cosx)
\(\frac{d}{dx}u(x) = \cos x\) \(\frac{d}{dx}v(x) = 1-\sin x\)

By quotient rule
\[\frac{d}{dx}\frac{u(x)}{v(x)} = \frac{v(x)\frac{d}{dx}u(x) + u(x)\frac{d}{dx}v(x)}{[v(x)]^2}\]
You know u(x), v(x), \(\frac{d}{dx}u(x) \), and \(\frac{d}{dx}v(x) \).
Can you do it now?

Answer 10

isnt quotient rule with a - on top

Answer 11

My bad :(
It should be a minus sign!! I'm sorry!!

Answer 12

ok

Answer 13

what did you get

Answer 14

i got \[((x+\cos x)(\cos x)-(1+\sin x)(1-\sin x))/(x+\cos x)^2\]

Answer 15

Same here, but I think we can simplify it

Answer 16

(x+cosx)(cosx)−(1+sinx)(1−sinx)
Distribute / use the identity \((a-b)(a+b) = a^2 - b^2\)
\[x\cos c + cos^2x - (1-sin^2x)\]
This identity would be useful here: \(sin^2x+cos^2x =1\)

Answer 17

Sorry. It should be \[x cosx + cos^2x - (1-sin^2x)\]for the third line

Answer 18

ok so what would the answer be then

Answer 19

Simplify this: \(xcosx+cos^2x−(1−sin^2x)\), what do you get?

Answer 20

xcosx

Answer 21

Yes.
So,
\[\frac{(x+cosx)(cosx)−(1+sinx)(1−sinx)}{(x+cosx)^2}\]\[=\frac{x\cos x}{(x+cosx)^2}\]

Answer 22

so it equals xcosx

Answer 23

No

Answer 24

sorry never mind