We have the equation
sin x + cos y = 1 {\displaystyle \sin x+\cos y=1}
We want to use implicit differentiation to find d y d x {\displaystyle {\frac {dy}{dx}}} . We take the derivative of both sides:
cos x − sin y d y d x = 0. {\displaystyle \cos x-\sin y{\frac {dy}{dx}}=0.}
Now, we solve for d y d x {\displaystyle {\frac {dy}{dx}}} :
sin y d y d x = cos x d y d x = cos x sin y {\displaystyle {\begin{aligned}\sin y{\frac {dy}{dx}}=\cos x\\{\frac {dy}{dx}}={\frac {\cos x}{\sin y}}\end{aligned}}}
And now, we plug in x = π 2 {\displaystyle x={\frac {\pi }{2}}} and π 2 {\displaystyle {\frac {\pi }{2}}} . We know that cos ( π 2 ) = 0 {\displaystyle \cos \left({\frac {\pi }{2}}\right)=0} and sin ( π 2 ) = 1 {\displaystyle \sin \left({\frac {\pi }{2}}\right)=1} , so we get
d y d x = 0 {\displaystyle {\frac {dy}{dx}}=0}
Answer: d y d x = 0 {\displaystyle \color {blue}{\frac {dy}{dx}}=0}
. We take the derivative of both sides:
:
and 
. We know that 
and 
, so we get
