Question

derivative of f(x)= cotx/sinx
im kind of stuck using the quotient rule on this one...

Answer 1

Do you mean cos x / sin x? If so, then you can just use substitution cos x = t, sin x = n, then derive n/t.

Answer 2

ohh. im still lost either way, can u expand on that?

Answer 3

The quotient rule is for f'(x) = n/t = (n * t') - (t * n') / t²

Now use what I've read above, and try to fill this formula.

Answer 4

that i know, and i did it, but my book says the answer is (-1-cos^2x)/sin^3x i want to know how it got to that. (i mean clearly im having trouble remembering basic trig identities..)

Answer 5

is the numerator cot x or cos x? I don't know how to come to that answer either...

Answer 6

the numerator is cotx. ugh ive been sitting here for a while wondering how my textbook got that answer -_- It seems like it would be simple, and im just forgetting something

Answer 7

start with the derivative of cot x? I don't know what it is, you could google it.

Answer 8

f(x)= cotx/sinx
= cotx * csc x, since 1/sinx = cscx
d/dx(cot(x) csc(x))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cot(x) and v = csc(x):
= cot(x) (d/dx(csc(x)))+csc(x) (d/dx(cot(x)))
The derivative of cot(x) is -csc^2(x):
= cot(x) (d/dx(csc(x)))+csc(x) (-csc^2(x))
The derivative of csc(x) is -cot(x) csc(x):
= cot(x) (-cot(x) csc(x))-csc^3(x)
= -csc(x)(cot^2(x) + csc^2(x))
= -csc(x)(cos^2(x)/sin^2(x) + csc^2(x))
= -csc(x)(cos^2(x)csc^2(x) + csc^2(x))
= -csc^3(x)(cos^2(x) + 1)
= (-1 - cos^2(x))csc^3(x)
= (-1 - cos^2(x)/sin^3(x)

Ans. (-1 - cos^2(x))/sin^3(x)

Answer 9

Alternatively, f(x) = cot x / sin x = cos x /sin^2 x

Hence applying the quotient rule:
df/dx = (sin^2 x .(-sin x) - cos x. 2 sin x. cos x)/sin^4 x
= -sin x ( sin^2 x + 2cos^2 x)/sin^4 x
= -(1 + cos^2 x)/sin^3 x

Answer 10

Nice job James!