Who can differentiate??

Question

y=csc x * cot x

y=sin (2x) + cos2 x



AHHH!!! I hate this crap.. It sucks massivly.

Answer 1

Just apply the product rule:

d(csc x cot x)/dx
= d(csc x)/dx cot x + csc x d(cot x)/dx
= - csc x cot x cot x + csc x (-csc² x)
= -csc x cot² x - csc³ x

And that's it for the first one. For the second, apply the chain rule:

d(sin (2x))/dx + d(cos² x)/dx
cos (2x) d(2x)/dx + 2 cos x d(cos x)/dx
2 cos (2x) - 2 cos x sin x

Or simplifying:

2 cos (2x) - sin (2x)

And we are done.

Answer 2

dy/dx = csc(x)(-csc^2(x)) + cot(x)(-cot(x)csc(x))
= -csc^3(x) - cot^2(x)csc(x)
= -csc^(x)( csc^2(x) + cot^2(x) )
= -csc^(x)(1 + 2cot^2(x) ).

If y = sin(2x) + cos(2x), then:
dy/dx = 2cos(2x) - 2sin(2x)
= 2( cos(2x) - sin(2x) ).

If y = sin(2x) + cos^2(x), then:
dy/dx = 2cos(2x) + 2cos(x)(-sin(x))
= 2cos(2x) - sin(2x).

Answer 3

y=csc x * cot x
dy/dx=

Answer 4

y = cosec x * cot x

cosec x * - cosec^2 (x) + (-cosec x * cot x)
= -cosec^3(x) - cosec(x)*cot(x)

y = sin 2x + cos 2x

2 cos (2x) - 2 sin (2x)

Answer 5

dy/dx=2cos(2x)-2sin(2x) for the second one