Question

How do I simplify this

Answer 1

http://prntscr.com/52xffq

Answer 2

from the first to the second one

Answer 3

I don't understand if you are doing dy/dx and have Ys on both sides, where are your y primes ?

Answer 4

oh sorry this was an implicit differentiation. I got the answer same as the top one but I can't simplify it further to get the answer at the bottom

Answer 5

http://prntscr.com/52xhih the original question is to find dy/dx using implicit differentiation

Answer 6

\(\LARGE\color{black}{ y\tan(x+y)=4}\)

Product rule, with a chain rule for tangent.

Answer 7

dy/dx

\(\Large\color{black}{ \frac{dy}{dx}(y)\times \tan(x+y)~+~ y \times \frac{dy}{dx}\tan(x+y)=\frac{dy}{dx}(4)}\)

Answer 8

My internet sucks, sorry.

Answer 9

We know that deriv. of tanx is sec^2x

\(\Large\color{black}{ y'\times \tan(x+y)~+~ y \times \sec^2(x+y)\times \frac{dy}{dx}(x+y)=0}\)

Answer 10

I am applying chain for tangent, ....

Answer 11

yeah I actually did it and arrived at this http://prntscr.com/52xjnp but I just need help on simplifiying it to this http://prntscr.com/52xk0n

Answer 12

gosh my language sucks. Sorry it isn't my first language

Answer 13

No...

Answer 14

Lets start over, okay?

Answer 15

\(\Large\color{black}{ y \tan(x+y)=4}\)

Answer 16

How about I post my written solution here? I hope you can read it. My writing sucks too

Answer 17

You are doing some unnecessary stuff.

Answer 18

\(\LARGE\color{black}{ y\tan(x+y)=4}\)

Product rule, with a chain rule for tangent.

Answer 19

At first,

\(\Large\color{black}{ \frac{dy}{dx}\tan(x+y)=}\)

\(\Large\color{black}{ \sec^2(x+y) \times\frac{dy}{dx}(x+y)}\)

\(\Large\color{black}{ \sec^2(x+y) \times(\frac{dy}{dx}x+\frac{dy}{dx}y)}\)

\(\Large\color{black}{ \sec^2(x+y) \times(1+y')}\)

Answer 20

Now, do the product rule, knowing the dy/dx of tan(x+y)

Answer 21

\(\Large\color{blue}{ \frac{dy}{dx} \tan(y)=\sec^2(x+y) \times(1+y')}\)
is given. (we know this)

\(\Large\color{black}{ y\tan(x+y)=4}\)


\(\large\color{black}{ \frac{dy}{dx}(y)\times\tan(x+y)+(y)\times\frac{dy}{dx}\tan(x+y)=\frac{dy}{dx}4}\)

Answer 22

MY internet again... I am sorry.

Answer 23

I hope you can finish, if not, or if something doesn't make sense to you, please ask.

Answer 24

http://prntscr.com/52xnkc. This..

Answer 25

http://prntscr.com/52xnkc

Answer 26

yes

Answer 27

re-wrote sec and tan in terms of cosine,