Hi,
Q1. Find derivative of cos(x+y) + y = 0
could anyone give me steps and the answer for this please?
thanks,
moon
If y is a function of x, then :
$\displaystyle \cos(x+y)$ has to be differentiated following the chain rule.
$\displaystyle [\cos(u(x))]'=-u'(x) \sin(u(x))$
So the derivative of $\displaystyle \cos(x+y)=-(1+y') \sin(x+y)$
So the derivative of $\displaystyle \cos(x+y)+y=0$ is $\displaystyle -(1+y')\sin(x+y)+y'=0$