How do you differentiate $f(x)=1+x-cosx$ ?

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Answer 1 · 2016-12-16T17:48:31

$f'(x)=1+sinx$

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 $f(x)=1+x-cosx$ , then

$f'(x)=0+1+-(-sinx) = 1+sinx$

How do you differentiate f(x)=1+x-cosxf(x)=1+x-cosx?