Question

How do you find g '(x) from g(x)=sin x+(1/2)cot x ?

Answer 1

well, the derivative of sin(x) = cos(x)
and the derivative of cot(x) = -csc^2(x)
So, for your function
g(x) =sin (x)+(1/2)cot (x)
you just separate it as 2 functions, so, g(x)=h(x)+(1/2)i(x), where
h(x)=sin(x), and i(x)=cot(x)
So, in the end
g'(x)=h'(x)+(1/2)i'(x)
and g'(x)= cos(x) - (1/2) csc^2(x)

Answer 2

So you can just leave the 1/2 in there, you don't have to do anything with it?

Answer 3

1/2 is a constant, when they're multiplying you just leave'em there, for example
\[f(x)=3\cos(2x)\]
then
\[f'(x)=3*-sen(2x)*2=-6sen(2x)\] got it?

Answer 4

@Umangiasd So what if g(x)= x^3cosx-13x sinx-13cosx ?
How would you do that?

Answer 5

First, you must apply the product rule, that says
for
\[f(x) = g(x) h(x)\]
\[f'(x)=g'(x)h(x)+g(x)h'(x)\]
in that case you must apply the rule for x^3 cos(x) and for xsinx, try and tell me what do ya get

Answer 6

Something like this: [(3x^2)(cos x)+(-sin x)(x^3)] - [(13)(sin x) + (cos x)(13x)] - 13(-sin x)

Answer 7

Excellent

Answer 8

So is that the final answer or should I do some more simplification?

Answer 9

you can put some stuff together as
\[3x^2\cos(x) -x^3\sin(x) -13xcos(x)\] the 13sen(x) got cancelled

Answer 10

Thanks for your help!