Question

HELP!!





Differentiate,check my answer please!

Answer 1

\[y=\frac{xtanx}{secx+tanx} find \frac{dy}{dx}\]

Answer 2

I'm getting,
\[secx+tanx({xsec^{2}x+tanx+xtanx(secxtanx+tanx)}\]

Answer 3

@hba

Answer 4

use the formula for the derivative of (u/v)= (vdu-udv)/v^2

Answer 5

if we use that formula
y=xtanx/(sec x+tan x)
dy/dx=
=[(sec x+tan x)(x sec^2 x+tan x)-(x tan x)(sec x tan x+sec^2 x)]/(sec x +tan x)^2
continue further until you cancel out the denominator
the denominator would be sec x+tan x

Answer 6

its the same i wrote or different?

Answer 7

different
dydx=
=[(sec x+tan x)(x sec^2 x+tan x)-(x tan x)(sec x tan x+sec^2 x)]/(sec x +tan x)^2
=[(sec x+tan x)(x sec^2 x+tan x)-x(sec x tan^2 x+ sec^2 x tan x)]/(sec x +tan x)^2
=[(sec x+tan x)(x sec^2 x+tan x)-x(sec x tan x)(tan x+ sec x )]/(sec x +tan x)^2
now try to factor out or cancel out sec x+tanx