Differentiate w.r.t x x3
sinx/cosx
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Answer:
Let y=cosx.cos2x.cos3x
Taking logarithm on both the sides, we obtain
logy=log(cosx.cos2x.cos3x)=log(cosx)+log(cos2x)+log(cos3x)
Differentiating both sides with respect to x, we obtain
y
1
dx
dy
=
cosx
1
.
dx
d
(cosx)+
cos2x
1
.
dx
d
(cos2x)+
cos3x
1
.
dx
d
(cos3x)
⇒
dx
dy
=y[−
cosx
sinx
−
cos2x
sin2x
.
dx
d
(2x)−
cos3x
sin3x
.
dx
d
(3x)]
⇒
dx
dy
=−cosx.cos2x.cos3x[tanx+2tan2x+3tan3x]
Step-by-step explanation:
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