How do you differentiate #f(x)=1+x-cosx#?
1 Answer
Dec 16, 2016
#f'(x)=1+sinx#
Explanation:
We need the following rules for differentiation
#d/dx cosx = -sinx#
#d/dx c = 0# , where#c# is a constant;
#d/dx x^n = nx^(n-1)# , where#n# is a constant;
So if
#f'(x)=0+1+-(-sinx) = 1+sinx#
