Question

Find the derivative of $f\left(x\right)$ from the first principles, where $f\left(x\right)$ is
$\sin x+\cos x$

Answer 1

$f\left(x\right)=\sin x+\cos x$

We know,

$f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

Now, $f(x+h)=\sin (x+h)+\cos (x+h)$

Therefore,

$f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{\sin (x+h)+\cos (x+h)-(\sin x+\cos x)}{h}$

$=li{m}_{h\to 0}\frac{\sin x\cos h+\cos x\sin h+\cos x\cos h-\sin x\sin h-\sin x-\cos x}{h}$

$=li{m}_{h\to 0}\frac{\sin h(\cos x-\sin x)+\sin x(\cos h-1)+\cos x(\cos h-1)}{h}$

$=li{m}_{h\to 0}\frac{\sin h(\cos x-\sin x)}{h}+li{m}_{h\to 0}\frac{\sin x(\cos h-1)}{h}+li{m}_{h\to 0}\frac{\cos x(\cos h-1)}{h}$

$=(\cos x-\sin x)li{m}_{h\to 0}\frac{\sin h}{h}-\sin xli{m}_{h\to 0}\frac{1-\cos h}{h}-\cos xli{m}_{h\to 0}\frac{1-\cos h}{h}$

$=\cos x-\sin x-0-0$

$=\cos x-\sin x$

Hence, $f^{\prime }\left(x\right)=\cos x-\sin x$