Differentiate the functions with respect to x.
\(cos(\sqrt x)\)
Let f(x )= \(cos(\sqrt x)\)
Also, let u(x) = \(\sqrt x\)
And, v(t) = cos t
Then, vou(x) = v(u(x))
= v(\(\sqrt x\))
= cos x
= f(x)
Clearly, f is a composite function of two functions, u and v, such that
t = u(x) = \(\sqrt x\)
Then, \(\frac {dt}{dx}\)=\(\frac {d}{dx}\)(\(\sqrt x\)) = \(\frac {d}{dx}\)(\(x^\frac 12\))= \(\frac12 x^{-\frac 12}\) = \(\frac {1}{2\sqrt x}\)
And \(\frac {dv}{dt}\) = \(\frac {d}{dt}\)(cos t) = -sin t
=-sin(\(\sqrt x\))
By using chain rule, we obtain
\(\frac {dt}{dx}\) = \(\frac {dv}{dt}\) . \(\frac {dt}{dx}\)
=-sin(\(\sqrt x\)) . \(\frac {1}{2\sqrt x}\)
=-\(\frac {sin\ \sqrt x}{2√x}\)
Alternate Method:
\(\frac {d}{dx}\)\([cos (\sqrt x)]\) = -sin(\(\sqrt x\)) . \(\frac {d}{dx}\)\((\sqrt x)\)
= -sin(\(\sqrt x\)) . \(\frac {d}{dx}\)(\(x^{\frac 12}\))
= -sin\(\sqrt x\). \(\frac 12\)\(x^{-\frac 12}\)
= -\(\frac {sin\ \sqrt x}{2√x}\)
The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} 2+x, & \text{if } x \geq 0 \\ 2-x, & \text{if } x \leq 0 \end{cases} \] Then function \( f(x) \) at \( x=0 \) is: