What is the derivative of y= x⁵ + cotx
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Answer:
y=ln(cosecx−cotx)
Chain rule,
dx
dy
=
du
dy
⋅
dx
du
Here u=cosecx−cotx
⇒
dx
dy
=
cosecx−cotx
1
⋅
dx
d(cosecx−cotx)
⇒
dx
dy
=
cosecx−cotx
−cosecx.cotx+cosec
2
x
⇒
dx
dy
=
cosecx−cotx
−cosecx(cotx−cosecx)
⇒
dx
dy
=cosecx
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