Differentiate - Trig?
Posted in xn--kfs74mzzid01b.com edit by jack on January 7th, 2009
y= x^2sinxtanx
dy/dx= x^2{sin(x) sec^2(x) + tan(x)cos(x)} + 2x sin(x) tan(x)
=x.tan(x) {x[sec(x)+cos(x)] + 2sin(x)}
d/dx (uvw) = du/dx vw + u dv/dx w + uv dw/dx
Let u = x²; du/dx = 2x
v = sinx; dv/dx = cosx
w = tanx; dw/dx = sec²x
Then d/dx (x² sinx tanx) becomes
2x sinx tanx + x² cosx tanx + x² sinx sec²x
Alternatively you could let u = x² sinx and v = tanx and use the standard product rule. You would then need to use it again for the product x² sinx.
u = sin x . tan x ---------[2]
v = x^2-------[3]
From [2] =>
u = sin x . tan x
du/dx = sin x . (sec^2 (x)) + tan x . (cos x) ; product rule
du/dx = sin x . sec^2 (x) + tan x . cos x
du/dx = (sin x / cos^2 (x)) + (sin x / cos x) . cos x
*du/dx =( tan x / cos x) + sin x---------------[4]
from [3] =>
v = x^2
*dv/dx = 2x ----------[5]
lets substitute u,v from [2] and [3] to our main eq [1]
y = (v) . (u) ; differentiate for production
dy / dx = {(v) . [du/dx]} + {(u) . [dv/dx] } ;
Next substitute values from [2] [3] [4] & [5]
dy / dx = { x^2 .[ ( tan x / cos x) + sin x] } + { sin x . tan x . [2x] }
u can simplyfy the answer further if you want
The more the merrier!!
Is it :-
y = (x²) sin x tan x
OR
y = x^(2 sin x) tan x
OR
y = x^(2 sin x tan x)
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