Question

If tan(x+y)=x then dy/dx=
Answer: cos^2(x+y)-1
Please show how to get to this answer.

Answer 1

do u know chain rule ?

Answer 2

yes i do

Answer 3

so where are you stuck ?
could you start ?

Answer 4

I got this far...

Answer 5

\[\sec ^{2}(x+y)(1+\frac{ dy }{ dx })\]
\[\sec ^{2}(x+y)(1+\frac{ dy }{ dx })-1\]
and i was kinda confused where to go from there

Answer 6

our main aim is to isolate dy/dx

distribute sec^2 (x+y) inside brackets

Answer 7

ok so i have...
\[[\sec^2x+\sec^2y](1+\frac{ dy }{ dx })-1=0\]

Answer 8

should i put the 1 back on the other side and divide by the [sec^2...]?

Answer 9

that can't be done... sec(A+B) does not = secA +secB

i was talking about this
\(\sec^2 (x+y)(1+y') = \sec^2(x+y) + y' \sec^2 (x+y)\)

Answer 10

oh ok. i got it now. thank you :)

Answer 11

could you find dy/dx properly ?
and welcome ^_^

Answer 12

yes. i got it :)