As in the hint, we have
f ( x ) = 1 g ( x ) , {\displaystyle f(x)={\frac {1}{g(x)}},}
where g ( x ) = sin ( x ) + cos ( x ) {\displaystyle g(x)=\sin(x)+\cos(x)} . Using the chain rule (with f ( x ) = h ( g ( x ) ) {\displaystyle f(x)=h(g(x))} , where h ( x ) = 1 / x {\displaystyle h(x)=1/x} ),
f ′ ( x ) = − 1 ( g ( x ) ) 2 × g ′ ( x ) . {\displaystyle f'(x)={\frac {-1}{(g(x))^{2}}}\times g'(x).}
We have g ′ ( x ) = sin ′ ( x ) + cos ′ ( x ) = cos ( x ) − sin ( x ) . {\displaystyle g'(x)=\sin '(x)+\cos '(x)=\cos(x)-\sin(x).} Putting it all together,
f ′ ( x ) = − cos ( x ) + sin ( x ) ( sin ( x ) + cos ( x ) ) 2 . {\displaystyle f'(x)={\frac {-\cos(x)+\sin(x)}{(\sin(x)+\cos(x))^{2}}}.}
Finally we can evaluate f' at − π / 2 {\displaystyle -\pi /2} :
f ′ ( − π 2 ) = − cos ( − π 2 ) + sin ( − π 2 ) ( sin ( − π 2 ) + cos ( − π 2 ) ) 2 = 0 + ( − 1 ) ( − 1 + 0 ) 2 = − 1 {\displaystyle f'\left(-{\frac {\pi }{2}}\right)={\frac {-\cos \left(-{\frac {\pi }{2}}\right)+\sin \left(-{\frac {\pi }{2}}\right)}{\left(\sin \left(-{\frac {\pi }{2}}\right)+\cos \left(-{\frac {\pi }{2}}\right)\right)^{2}}}={\frac {0+(-1)}{\left(-1+0\right)^{2}}}=-1}
. Using the chain rule (with 
, where 
),
Putting it all together,
:
