2006-11-01 15:59:13 · 6 answers · asked by soccerman2431 1 in Science & Mathematics ➔ Mathematics
1=sec^2(x+y)(1+y')
so, y'=cos^2(x+y)-1
2006-11-01 16:02:31 · answer #1 · answered by askance 4 · 0⤊ 0⤋
dy/dx = 1+tan(x+y)^2
2006-11-02 00:00:23 · answer #2 · answered by Guru 6 · 0⤊ 0⤋
Y=1
x= -1
2006-11-02 00:00:51 · answer #3 · answered by Anonymous · 0⤊ 1⤋
i dun know whether this's right or wrong...
i just do it on my own...
1st, using the division rule : [v(d/dx0u-u(d/dx)v]/v^2
then...
u do the nonsence... n u'll get
=1/cos(x+y)^2.[cos(x+y)dy/dx(cos(x+y)+sin(x+y)dy/dxsin(x+y)]
by factoring the dy/dx,
=1/cos(x+y)^2.[(dy/dx)(cos^2(x+y)+sin^2(x+y))]
we know that,
sin^a+cos^a =1
and,
1/cos^2a=sec^2a,
hence,
simplify the equation, we'll get
=sec^2(x+y)dy/dx
TA-DA!!! ^_^
2006-11-02 01:27:19 · answer #4 · answered by azfarbeckham 1 · 0⤊ 0⤋
y = atan(x) - x
dy/dx = 1/(1+x^2) - 1
2006-11-02 00:10:24 · answer #5 · answered by Helmut 7 · 0⤊ 1⤋
dy/dx=-sin²(x+y)
2006-11-02 00:03:52 · answer #6 · answered by venomfx 4 · 0⤊ 0⤋