Derivatives — MATHS STD 11 Science — Question

$\text{f}(\text{x})=\sin\text{x}+\cos\text{x}$

$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\sin(\text{x}+\text{h})+\cos(\text{x}+\text{h})\big\}-\sin\text{x}+\cos\text{x}}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\sin(\text{x}+\text{h})+\cos(\text{x}+\text{h})-\sin\text{x}-\cos\text{x}\big\}}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\sin(\text{x}+\text{h})-\sin\text{x}\big\}+\big\{\cos(\text{x}+\text{h})-\cos\text{x}\big\}}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big\{2\sin\frac{(\text{x}+\text{h}-\text{x})}{2}\cos\frac{(\text{x}+\text{h}+\text{x})}{2}\Big\}+\Big\{-2\sin\frac{\text{x}+\text{h}+\text{x}}{2}\sin\frac{\text{x}+\text{h}-\text{x}}{2}\Big\}}{\text{h}}$

$\begin{bmatrix}\because\sin\text{A}-\sin\text{B}=2\sin\frac{\text{A}-\text{B}}{2}\cos\frac{\text{A}+\text{B}}{2}\\\cos\text{A}-\cos\text{B}=2\sin\frac{\text{A}+\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}{}\end{bmatrix}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}.\cos\frac{\text{2x}+\text{h}}{2}-2\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)\sin\text{h}}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\Big\{\cos\frac{\text{x}+\text{h}}{2}-\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)\Big\}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$

$=\cos\text{x}-\sin\text{x}$