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\(\sqrt x\). \(\frac 12\)\(x^{-\frac 12}\)
= -\(\frac {sin\ \sqrt x}{2√x}\)
Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}