Question

Use the first principle of differentiation to differentiate $y = \sqrt {\sin x} $.

Answer 1

Hint: The given question requires us to find the derivative of a function using the first principle of differentiation. The first principle of differentiation helps us evaluate the derivative of a function using limits. Calculating the derivative of a function using the first principle of differentiation may be a tedious task. We may employ identities and tricks to calculate the limits and evaluate the required derivative.

Complete step by step solution:
We have to evaluate the derivative of $y =f(x) = \sqrt {\sin x} $ using the first principle of differentiation.
According to the first principle of differentiation, the derivative of a function can be evaluated by calculating the limit \[f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f{\text{(x)}}}}{h}\] .
So, the derivative of the function $y = \sqrt {\sin x} $ can be calculated by the first rule of differentiation as:
$f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sqrt {\sin \left( {x + h} \right)} - \sqrt {\sin x} }}{h}} \right]$
Expanding the sine of a compound angle using the compound angle formula for sine as $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, we get,
$ \Rightarrow f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sqrt {\sin x\cosh + \cos x\sinh } - \sqrt {\sin x} }}{h}} \right]$
Multiplying and dividing the rational expression by $\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)$, we get,
$ \Rightarrow f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sqrt {\sin x\cosh + \cos x\sinh } - \sqrt {\sin x} }}{h}} \right] \times \dfrac{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}$
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get,
$ \Rightarrow f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sin x\cosh + \cos x\sinh - \sin x}}{{h\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Taking the common terms outside the bracket, we get,
$ \Rightarrow f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sin x\left( {\cosh - 1} \right) + \cos x\sinh }}{{h\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Separating out the denominator, we get,
$ \Rightarrow f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\dfrac{{\sin x\left( {\cosh - 1} \right)}}{h} + \dfrac{{\cos x\sinh }}{h}}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Using the half angle formula for cosine in numerator,
$ \Rightarrow f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\dfrac{{\sin x\left( {1 - 2{{\sin }^2}\dfrac{h}{2} - 1} \right)}}{h} + \dfrac{{\cos x\sinh }}{h}}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Cancelling the like terms with opposite signs,
$ \Rightarrow f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{ - 2\sin x\dfrac{{\left( {{{\sin }^2}\dfrac{h}{2}} \right)}}{h} + \cos x\left( {\dfrac{{\sinh }}{h}} \right)}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Separating out the limits for each expression, we get,
$ \Rightarrow f'(x){\text{ = }}\left[ {\mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2\sin x\dfrac{{\left( {{{\sin }^2}\dfrac{h}{2}} \right)}}{h}}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}} + \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x\left( {\dfrac{{\sinh }}{h}} \right)}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Using the standard limit result $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$, we get,
$ \Rightarrow f'(x){\text{ = }}\left[ {\mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2\sin x\dfrac{{\left( {{{\sin }^2}\dfrac{h}{2}} \right)}}{h}}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}} + \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x\left( 1 \right)}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Doing some modifications in the first term,
\[ \Rightarrow f'(x){\text{ = }}\left[ {\mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2\sin x\dfrac{{\left( {{{\sin }^2}\dfrac{h}{2}} \right)}}{{\dfrac{4}{h} \times {{\left( {\dfrac{h}{2}} \right)}^2}}}}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}} + \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x\left( 1 \right)}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]\]
$ \Rightarrow f'(x){\text{ = }}\left[ {\mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2\sin x \times \dfrac{h}{4}}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}} + \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x\left( 1 \right)}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
$ \Rightarrow f'(x){\text{ = }}\left[ {\mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2h\sin x}}{{4\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}} + \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x}}{{\left( {\sqrt {\sin x\cosh + \cos x\sinh } + \sqrt {\sin x} } \right)}}} \right]$
Putting in the value of limits, we get,
$ \Rightarrow f'(x){\text{ = }}\left[ {\dfrac{{ - 2\left( 0 \right)\sin x}}{{4\left( {\sqrt {\sin x\cos 0 + \cos x\sin 0} + \sqrt {\sin x} } \right)}} + \dfrac{{\cos x}}{{\left( {\sqrt {\sin x\cos 0 + \cos x\sin 0} + \sqrt {\sin x} } \right)}}} \right]$
Simplifying the expression,
$ \Rightarrow f'(x){\text{ = }}\left[ {0 + \dfrac{{\cos x}}{{\left( {\sqrt {\sin x\left( 1 \right) + \cos x\left( 0 \right)} + \sqrt {\sin x} } \right)}}} \right]$
$ \Rightarrow f'(x){\text{ = }}\dfrac{{\cos x}}{{2\sqrt {\sin x} }}$
Therefore, the derivative of the function $y = \sqrt {\sin x} $ is $f'(x){\text{ = }}\dfrac{{\cos x}}{{2\sqrt {\sin x} }}$.

Note:
The derivative of the given function can also be calculated by using the chain rule and power rule of differentiation. According to the power rule of differentiation, the derivative of ${x^n}$ is $n{x^{n - 1}}$ .
So, going by the chain rule of differentiation, the derivative of $y = \sqrt {\sin x} $ is $\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {\sin x} }} \times \left( {\cos x} \right)$. So, the derivative of the given function is $f'(x){\text{ = }}\dfrac{{\cos x}}{{2\sqrt {\sin x} }}$ .