To determine where the function
f(x)=tan−1(sin‌x+cos‌x) is increasing, we need to analyze its derivative. A function is increasing where its derivative is positive.
First, let's find the derivative of
f(x). Given
f(x)=tan−1(sin‌x+cos‌x), we use the chain rule:
Let
u=sin‌x+cos‌x. Then,
f(x)=tan−1(u) and
‌[tan−1(u)]=‌‌Next, we compute
‌ :
u=sin‌x+cos‌x So,
‌=cos‌x−sin‌xThus:
f′(x)=‌(cos‌x−sin‌x)To find where
f′(x)>0, examine the expression:
cos‌x−sin‌x>0Rewriting this, we have:
cos‌x>sin‌x To identify the intervals where
cos‌x>sin‌x, let's take a look at the behavior of the basic trigonometric functions. We know that:
cos‌x=sin‌x when
x=‌+kπ for any integer
k.
In the interval
(−‌,‌),cos‌x is generally greater than
sin‌x.
Verifying the intervals provided, we conclude that
cos‌x>sin‌x holds true for
x∈(−‌,‌).
Hence, the function
f(x)=tan−1(sin‌x+cos‌x) is an increasing function in:
Option C:
(−‌,‌).