Question

DE

Answer 1

\[y=\frac{xsinx}{sinx+cosx}\]

Answer 2

Find dy/dx

Answer 3

I think you should try to differentiate it your self

Remember,

(U/V) formula.

Answer 4

\[\frac{(sinx+cosx)d(xsinx)-xsinxd(sinx+cosx)}{(sinx+cosx)^{2}}\]

Answer 5

that is /dx is that so?

Answer 6

is it -xsinx?

Answer 7

Yes, that's the one.

one trick in answering these kinds of questions is using natural logs to simplify the thing before you proceed.

\(\ln y= \ln x + \ln sinx - \ln(\sin x+\cos x)\)

this is much easier on the eyes...lol

Answer 8

wait i guess ive done it wrong :X

Answer 9

and i dont really know logs

Answer 10

ah well, then just use the original formula then. Godspeed.
though \( (\ln u)'=\frac{u'}{u}\), this is the one using log.

Answer 11

CAN U PLEASE TELL ME THE NEXT STEP OF WHAT I WROTE ABOVE :/

Answer 12

are you familiar with the chain rule? you'll need it for \(d(x sin x)\)

Answer 13

yes I am,just write the next step so i can confirm it!!

Answer 14

i solved d(xsinx) as
(1+cosx)

Answer 15

by the product rule,
d(xsinx)
=x d(sinx) + sin x dx
=x cos x+ sinx
though

Answer 16

okay yeah!

Answer 17

so the whole complex thing becomes:
\[\frac{(sin x + cos x)(x \cos x + sin x)-x \sin x(cos x - sin x)}{(sin x + cos x)^2}\].
note that some terms will cancel out beautifully...lol

Answer 18

\[\frac{(xcosx+sinx)-sinx(cosx-sinx)}{(sinx+cosx)} \]
how far do i simplify after this?

Answer 19

@shadowys

Answer 20

how did you get that? no, you can't cancel one side only. Both of them on the denominator must have (sinx + cos x) to cancel out. Expand the original one.

Answer 21

IM GETTING TANX/2 O_O

Answer 22

oh my. lol can I see your steps? in a pic perhaps?

Answer 23

\[\frac{(xcosxsinx+\sin^{2]x+xcos^{2}x-xsinxcosx+xsin^{2}x}}{(sinx+cosx)^{2}} \]

Answer 24

what happned to the text lol

Answer 25

\[\frac{xcosxsinx+\sin^{2}x+xcos^{2}x-xsinxcosx+xsin^{2}x}{(sinx+cosx^{2})} \]

Answer 26

something like this?

Answer 27

so \[\frac{1+\sin^{2}x}{(1+2sinxcosx)}\]

Answer 28

i think you missed a sin x cos x up there, from (sinx+cosx)(xcosx+sinx)

Answer 29

ohyeah :p

Answer 30

numerator==
1+sinxcosx+sin^2x ?

Answer 31

yeah, but \(x cos^2 x +x sin^2 x\)= x, too,~

Answer 32

AH
so
x+sinxcosx+sin^2x/1+2sinxcosx?!

Answer 33

yh, can't be simplified further, i guess

Answer 34

:D!