How do you use the chain rule to differentiate $f(x)=sin(1/(x^2+1))$ ?

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Answer 1 · 2018-03-26T11:45:30

See below

$f(x)=Sin(1/(x^2+1))$

$f(x)=Sin(u)$

$f'(x)=Cos(u) times u'$

$u=1/(x^2+1)=(x^2+1)^-1$

(Substitute again to include the inner function)

$v=x^2+1$
$v'=2x$

$u'=-1(v')(v)^-2=-1(2x)(x^2+1)^-2=(-2x)/(x^2+1)^2$

Hence:

$f'(x)=((-2x)/(x^2+1)^2)Cos(1/(x^2+1))$

How do you use the chain rule to differentiate f(x)=sin(1/(x^2+1))f(x)=sin(1/(x^2+1))?