Use the first principle to differentiate? y=sqrt(sinx)

Mar 18, 2018

Step one is to rewrite the function as a rational exponent $f \left(x\right) = \sin {\left(x\right)}^{\frac{1}{2}}$

Explanation:

After you have your expression in that form, you can differentiate it using the Chain Rule:

In your case: ${u}^{\frac{1}{2}} \to \frac{1}{2} S \in {\left(x\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} S \in \left(x\right)$

Then, $\frac{1}{2} S \in {\left(x\right)}^{- \frac{1}{2}} \cdot C o s \left(x\right)$ which is your answer

Mar 18, 2018

$\frac{d}{\mathrm{dx}} \sqrt{\sin x} = \cos \frac{x}{2 \sqrt{\sin x}}$

Explanation:

Using the limit definition of the derivative we have:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

So for the given function, where $f \left(x\right) = \sqrt{\sin x}$, we have:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sqrt{\sin \left(x + h\right)} - \sqrt{\sin x}}{h}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sqrt{\sin \left(x + h\right)} - \sqrt{\sin x}}{h} \cdot \frac{\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}}{\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin \left(x + h\right) - \sin x}{h \left(\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}\right)}$

Then we can use the trigonometric identity:

$\sin \left(A + B\right) \equiv \sin A \cos B + \cos A \sin B$

Giving us:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h \left(\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}\right)}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin x \left(\cos h - 1\right) + \cos x \sin h}{h \left(\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}\right)}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin x \left(\cos h - 1\right)}{h \left(\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}\right)} + \frac{\cos x \sin h}{h \left(\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}\right)}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\cos h - 1}{h} \frac{\sin x}{\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}} + \frac{\sin h}{h} \frac{\cos x}{\sqrt{\sin \left(x + h\right)} + \sqrt{\sin x}}$

Then we use two very standard calculus limits:

${\lim}_{\theta \to 0} \sin \frac{\theta}{\theta} = 1$, and ${\lim}_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0$, and 

And we can now evaluate the limits:

$f ' \left(x\right) = 0 \times \frac{\sin x}{\sqrt{\sin \left(x\right)} + \sqrt{\sin x}} + 1 \times \frac{\cos x}{\sqrt{\sin \left(x\right)} + \sqrt{\sin x}}$

 \ \ \ \ \ \ \ \ \ = (cosx) / (2sqrt(sin(x)) #