f(x) = 1 - cos x / x^2 + 1
Quotient rule.
f(x) = 1 - cos x
f'(x) = 1 + sin x
g(x) = x^2 + 1
g'(x) = 2x
1 + sin x (x^2 + 1) - 2x(1-cosx) / (x^2 + 1 )^2
Am I correct?

Question

f(x) = 1 - cos x / x^2 + 1 
Quotient rule.
f(x) = 1 - cos x
f'(x) = 1 + sin x
g(x) = x^2 + 1
g'(x) = 2x 
1 + sin x (x^2 + 1) - 2x(1-cosx) / (x^2 + 1 )^2
Am I  correct?

shamil98 · Answer

oops, made a mistake..

luigi0210 · Answer

Yup

shamil98 · Answer

sin x (x^2 + 1) - 2x(1-cosx) / (x^2 + 1 )^2
there is no 1 +..

anonymous · Answer

Not Quite f'(x) should just be Sin(x)

anonymous · Answer

But otherwise that is right.

shamil98 · Answer

thanks.

anonymous · Answer

$$f'\left( \frac{ u }{ v } \right)=\frac{ vu'-uv' }{v ^{2} }$$

anonymous · Answer

$$f'\left( \frac{ 1-\cos x }{ x ^{2}+1 } \right)=\frac{ \sin x \left( x ^{2}+1 \right)-2x \left( 1-\cos x \right) }{\left(  x ^{2} +1 \right)^{2} }$$
you are absolutely correct.