Question

Derivative of ytan(x+y)=4 ?

Answer 1

derivative of 4 is 0

for the other side its a bit trickier

you need to recall the product rule

and also the chain rule

\[[y(\tan(x+y)]'=y'\tan(x+y)+y[\tan(x+y)]'\]

now need to find

\[[\tan(x+y)]'\]
which needs chain rule

\[[\tan(x+y)]'=(x+y)'\sec^2(x+y)=(1+y')\sec^2(x+y)\]

Answer 2

\[\frac{dy}{dx}\times \tan(x+y) +y \times \sec^2{(x+y)} \left( 1 + \frac{dy}{dx}\right)\]

Answer 3

oh wow

Answer 4

so you have

\[y' \tan(x+y)+y (1+y')\sec^2(x+y)=0\]

Answer 5

Thanks so much- any way to get dy/dx on one side by itself?

Answer 6

yes thats what you do is solve for y'

Answer 7

do some distribution
put your terms that have y' together and everything without the factor y' on the other side

then factor the y' out of the terms that did have y' in common and divide on both sides by whatever is being multiplied by y'

Answer 8

So after distributing:\[dy/dx * \tan (x+y) + y(\sec ^{2}(x+y)+ dy/dx * \sec ^{2}(x+y)\]
Correct?

Answer 9

Pre- solving, post- distribution.

Answer 10

ok so the close parenthesis is meant to be at the end if so then so far good
also distribute y to both of those terms inside that parenthesis ( and i think you meant to close at the very end)

Answer 11

\[dy/dx∗\tan(x+y)+(ysec2(x+y)+dy/dx∗ysec2(x+y))\]
This? Then solve for dy/dx?

Answer 12

Those are sec^2, not sec 2(x+y).

Answer 13

ok good

Answer 14

now put your terms that have y' together and put everything else on the other side of equation

Answer 15

\[\left(\begin{matrix}dy \\ dx\end{matrix}\right) \tan (x+y) + \left(\begin{matrix}dy \\ dx\end{matrix}\right) y \sec ^{2}(x+y) = -y \sec ^{2}(x+y)\]
Factor out dy/dx and divide by what is then (tan(x+y) + y sec^2(x+y)), leaving
\[\left(\begin{matrix}dy \\ dx\end{matrix}\right) = \left(\begin{matrix}(-y \sec ^{2}(x+y)) \\ (\tan (x+y) + y \sec ^{2}(x+y))\end{matrix}\right)\]
Is that the answer?

Answer 16

looks awesome

Answer 17

Woohoo! Thank you so much, I really appreciate it.