In this derivative problem, it is given that $y = \sqrt{\sin{x}}$. So, the differentiation of square root of $\sin{x}$ with respect to $x$ have to calculated to find the derivative of $y$ with respect to $x$. It can be done in two different methods.
We know that $y = \sqrt{\sin{x}}$ and take derivative both sides for finding the differentiation of $y$ with respect to $x$.
$\implies$ $\dfrac{d}{dx}{\, y}$ $\,=\,$ $\dfrac{d}{dx}{\, \sqrt{\sin{x}}}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\, \sqrt{\sin{x}}}$
In differential calculus, there is no direct formula to find the derivative of square root of $\sin{x}$ function. The $\sqrt{\sin{x}}$ function is a composition of two functions $\sqrt{x}$ and $\sin{x}$. Hence, the differentiation of $\sqrt{\sin{x}}$ can be calculated by the chain rule.
Take $u = \sin{x}$, then $\dfrac{d}{dx}{\, u} = \dfrac{d}{dx}{\, \sin{x}}$ as per derivative of sinx formula. Therefore, $\dfrac{du}{dx} = \cos{x}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\, \sqrt{u}}$
The right hand side of the equation can be written as follows as per chain rule.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{du}{dx} \times \dfrac{d}{du}{\, \sqrt{u}}$
It is calculated about that $\dfrac{du}{dx} = \cos{x}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\cos{x} \times \dfrac{d}{du}{\, \sqrt{u}}$
Now, find the derivative of square root of $u$ with respect to $u$ by the derivative of square root formula.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\cos{x} \times \dfrac{1}{2\sqrt{u}}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\cos{x}}{2\sqrt{u}}$
Actually, $u = \sin{x}$ in this example. So, replace the $u$ by its actual value.
$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\cos{x}}{2\sqrt{\sin{x}}}$
The derivative of square root of $\sin{x}$ with respect to $x$ can be calculated from first principle. According to the definition of the derivative, the differentiation of $\sqrt{\sin{x}}$ can be written in limit form.
Take $y = f{(x)}$. So, $f{(x)} = \sqrt{\sin{x}}$, then $f{(x+\Delta x)} = \sqrt{\sin{(x+\Delta x)}}$
$\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f{(x+\Delta x)}-f{(x)}}{\Delta x}}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f{(x+\Delta x)}-f{(x)}}{\Delta x}}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\sqrt{\sin{(x+\Delta x)}}-\sqrt{\sin{x}}}{\Delta x}}$
Take $h = \Delta x$ and express the expression in terms of $h$.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sqrt{\sin{(x+h)}}-\sqrt{\sin{x}}}{h}}$
Let’s try to calculate differentiation of square root of $\sin{x}$ function by evaluating the limit of trigonometric function by direct substitution method.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\sqrt{\sin{(x+0)}}-\sqrt{\sin{x}}}{0}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\sqrt{\sin{x}}-\sqrt{\sin{x}}}{0}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{0}{0}$
it is calculated that the derivative of the $\sqrt{\sin{x}}$ with respect $x$ is indeterminate. So, it is not possible to find the derivative by evaluating limit by the direct substitution approach.
Now, try to evaluate limit of the trigonometric function by the rationalization method to find the derivative of $\sqrt{\sin{x}}$ function with respect to $x$.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sqrt{\sin{(x+h)}}-\sqrt{\sin{x}}}{h}}$ $\times$ $1\Bigg]$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sqrt{\sin{(x+h)}}-\sqrt{\sin{x}}}{h}}$ $\times$ $\dfrac{\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}}{\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}}\Bigg]$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\Big(\sqrt{\sin{(x+h)}}-\sqrt{\sin{x}}\Big) \times \Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)}{h \times {\Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)}} \Bigg]}$
The numerator represents the special product of binomials and their product can be written as per difference of squares formula.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{{\Big(\sqrt{\sin{(x+h)}}\Big)}^2-{\Big(\sqrt{\sin{x}}\Big)}^2}{h \times {\Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)}} \Bigg]}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin{(x+h)}-\sin{x}}{h \times {\Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)}} \Bigg]}$
Use difference to product transformation trigonometric identity to combine the sine functions in the numerator.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{2\cos{\Bigg(\dfrac{x+h+x}{2}\Bigg)}\sin{\Bigg(\dfrac{x+h-x}{2}\Bigg)}}{h \times {\Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)}} \Bigg]}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{2\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)}\sin{\Bigg(\dfrac{\cancel{x}+h-\cancel{x}}{2}\Bigg)}}{h \times {\Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)}} \Bigg]}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{2\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)}\sin{\Bigg(\dfrac{h}{2}\Bigg)}}{h \times {\Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)}} \Bigg]}$
Now, factorize the function as product of two trigonometric functions as follows.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)} \times 2\sin{\Bigg(\dfrac{h}{2}\Bigg)}}{{\Big(\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}\Big)} \times h} \Bigg]}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\Bigg(\dfrac{\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)}}{\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}}\Bigg) \times \Bigg(\dfrac{2\sin{\Bigg(\dfrac{h}{2}\Bigg)}}{h}\Bigg)\Bigg]}$
Now, use product rule of limits to find the limit of product by the product of their limits.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)}}{\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin{\Bigg(\dfrac{h}{2}\Bigg)}}{h}\Bigg)}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)}}{\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg(\dfrac{h}{2}\Bigg)}}{\dfrac{h}{2}}\Bigg)}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)}}{\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\frac{h}{2} \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg(\dfrac{h}{2}\Bigg)}}{\dfrac{h}{2}}\Bigg)}$
Take $q = \dfrac{h}{2}$ and replace $\dfrac{h}{2}$ by $q$ only in second factor.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{\Bigg(\dfrac{2x+h}{2}\Bigg)}}{\sqrt{\sin{(x+h)}}+\sqrt{\sin{x}}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\sin{q}}{q}\Bigg)}$
Find the value of first factor by evaluating limit of the trigonometric function by the direct substitution method and then find value of the second factor by limit of sinx/x as x approaches 0 standard form formula.
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\Bigg(\dfrac{\cos{\Bigg(\dfrac{2x+0}{2}\Bigg)}}{\sqrt{\sin{(x+0)}}+\sqrt{\sin{x}}}\Bigg)$ $\times$ $\Bigg(1\Bigg)$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\cos{\Bigg(\dfrac{2x}{2}\Bigg)}}{\sqrt{\sin{x}}+\sqrt{\sin{x}}}$
$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\require{cancel} \dfrac{\cos{\Bigg(\dfrac{\cancel{2}x}{\cancel{2}}\Bigg)}}{2\sqrt{\sin{x}}}$
$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\cos{x}}{2\sqrt{\sin{x}}}$
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