Differentiate the following functions from first principles:

We have to find the derivative of $e^{\cos x}$ with the first principle method, so,

$f(x)=e^{\cos x}$

by using the first principle formula, we get,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\cos (x+h)}-e^{\cos x}}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\cos x}\left(e^{\cos (x+h)-\cos x}-1\right)}{h}$

[By using $\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1$ ]

$f^{\prime}(x)=\lim _{h \rightarrow 0} e^{\cos x} \frac{\cos (x+h)-\cos x}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} e^{\cos x} \frac{\cos x \cosh -\sin x \sin h-\cos x}{h}$

[By using $\cos (x+h)=\cos x \cosh -\sin x \sinh$ ]

$f^{\prime}(x)=\lim _{h \rightarrow 0} e^{\cos x}\left[\frac{\cos x(\cosh -1)}{h}-\frac{\sin x \sinh }{h}\right]$

[By using $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ and

$\left.\cos 2 x=1-2 \sin ^{2} x\right]$

$f^{\prime}(x)=\lim _{h \rightarrow 0} e^{\cos x}\left[\frac{\cos x\left(-2 \sin ^{2} \frac{h}{2}\right)\left(\frac{h}{4}\right)}{h\left(\frac{h}{4}\right)}-\sin x\right]$

$f^{\prime}(x)=\lim _{h \rightarrow 0} e^{\cos x}\left[\frac{\cos x\left(-2 \sin ^{2} \frac{h}{2}\right)\left(\frac{h}{4}\right)}{\frac{h^{2}}{2^{2}}}-\sin x\right]$

$f^{\prime}(x)=-e^{\cos x} \sin x$