Question

Differentiate from first principle:

$\left(vi\right)\sin x+\cos x$

Answer 1

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

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ in above expression, we get:

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

Applying the formula,

$\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$\cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2\cos \left(\frac{x+h+x}{2}\right)\sin \left(\frac{x+h-x}{2}\right)-2\sin \left(\frac{x+h+x}{2}\right)\sin \left(\frac{x+h-x}{2}\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=2\underset{h\to 0}{lim}\frac{\cos (x+\frac{h}{2})\sin \left(\frac{h}{2}\right)-\sin (x+\frac{h}{2})\sin \left(\frac{h}{2}\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{[\cos (x+\frac{h}{2})-\sin (x+\frac{h}{2})]\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$

$\Rightarrow f^{\prime }\left(x\right)=\cos (x+0)-\sin (x+0)[\because \underset{h\to 0}{lim}\frac{\sin h}{h}=1]$

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

Therefore, the derivative of $\sin x+\cos x\text{is}$
$\cos x-\sin x.$