Find the derivative of (sinx + cosx) from first principle.

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Answer

We have, $f(x)=\sin x+\cos x$
By using first principle of derivative
$\begin{array}{l}f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sin (x+h)+\cos (x+h)-\sin x-\cos x}{h} \\ =\lim _{h \rightarrow 0} \frac{[\sin x \cdot \cos h+\cos x \cdot \sin h+\cos x \cdot \cos h-\sin x \cdot \sin h-\sin x-\cos x]}{h}\end{array}$
$(\because \sin (x+y)=\sin x \cos y+\cos x \sin y$ and $\cos (x+y)=\cos x \cos y-\sin x \sin y)$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{\sin h(\cos x-\sin x)+\sin x(\cos h-1)+\cos x(\cos h-1)}{h} \\ =\lim _{h \rightarrow 0} \frac{\sin h}{h}(\cos x-\sin x)+\lim _{h \rightarrow 0} \frac{\sin x(\cos h-1)}{h}+\lim _{h \rightarrow 0} \frac{\cos x(\cos h-1)}{h} \\ =1 \cdot(\cos x-\sin x)+\lim _{h \rightarrow 0} \sin x\left[\frac{-(1-\cos h)}{h}\right]+\lim _{h \rightarrow 0} \cos x\left[\frac{-(1-\cos h)}{h}\right]\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ =(\cos x-\sin x)-\sin x \cdot \lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h}\right)-\cos x \cdot \lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h}\right) \\ =(\cos x-\sin x)-\sin x \cdot \lim _{h \rightarrow 0} \frac{2 \sin ^2 \frac{h}{2}}{h \times \frac{h}{4}} \times \frac{h}{4}-\cos x \cdot \lim _{h \rightarrow 0} \frac{2 \sin ^2 \frac{h}{2}}{h \times \frac{h}{4}} \times \frac{h}{4} \\ =(\cos x-\sin x)-\sin x \cdot 2 \cdot \frac{1}{4} \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^2 \times h-\cos x \cdot 2 \cdot \frac{1}{4} \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^2 h \\ =(\cos x-\sin x)-\frac{1}{2} \cdot \sin x \cdot(1) \times 0-\cos x \cdot \frac{1}{2} \cdot(1) \times 0\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]\end{array}$
= (cos x - sin x) - 0 - 0
= cos x - sin x