Derivative of $ \sqrt x + sinx $

Question

Please help me find out derivative of $\sqrt x + \sin x$. This is to be solved using first principle of derivatives only.

Answer 1

$$\begin{array}{rcl} \displaystyle \frac {\mathrm d} {\mathrm dx} \sqrt x + \sin x &=& \displaystyle \lim_{h \to 0} \frac {(\sqrt{x+h}+\sin(x+h)) - (\sqrt x+\sin x)} {h} \\ &=& \displaystyle \lim_{h \to 0} \frac {\sqrt{x+h}-\sqrt x} h + \frac {\sin(x+h)-\sin x} h \\ &=& \displaystyle \lim_{h \to 0} \frac {(\sqrt{x+h}-\sqrt x) (\sqrt{x+h}+\sqrt x)} {h (\sqrt{x+h}+\sqrt x)} + \frac {2\cos(x+h/2)\sin(h/2)} h \\ &=& \displaystyle \lim_{h \to 0} \frac {(x+h) - x} {h (\sqrt{x+h}+\sqrt x)} + \cos(x+h/2) \frac {\sin(h/2)} {h/2} \\ &=& \displaystyle \lim_{h \to 0} \frac h {h (\sqrt{x+h}+\sqrt x)} + \cos(x+h/2) \frac {\sin(h/2)} {h/2} \\ &=& \displaystyle \lim_{h \to 0} \frac 1 {\sqrt{x+h}+\sqrt x} + \cos(x+h/2) \frac {\sin(h/2)} {h/2} \\ &=& \displaystyle \frac 1 {\sqrt{x}+\sqrt x} + \cos(x) \\ &=& \displaystyle \frac 1 {2\sqrt{x}} + \cos(x) \\ \end{array}$$