Question

Differentiate the following functions from first principles:

e^{cos x}

Answer 1

$$\begin{gathered} \mathrm{Let}f\left(x\right)=e^{\cos x} \\ \Rightarrow f\left(x+h\right)=e^{\cos \left(x+h\right)} \\ \therefore \frac{d}{dx}\left\{f\left(x\right)\right\}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{\cos \left(x+h\right)}-e^{\cos x}}{h} \\ =\underset{h\to 0}{\lim }{e}^{\cos x}\left[\frac{e^{\cos \left(x+h\right)-\cos x}-1}{h}\right] \\ =\underset{h\to 0}{\lim }{e}^{\cos x}\left[\frac{e^{\cos \left(x+h\right)-\cos x}-1}{\cos \left(x+h\right)\cos x}\right]\times \frac{\cos \left(x+h\right)-\cos x}{h} \\ =\underset{h\to 0}{e^{\cos x}\lim }\left(\frac{\cos \left(x+h\right)-\cos x}{h}\right)\times \underset{h\to 0}{\lim }\left[\frac{e^{\cos \left(x+h\right)-\cos x}-1}{\cos \left(x+h\right)-\cos x}\right] \\ =e^{\cos x}\underset{h\to 0}{\lim }\left(\frac{\cos \left(x+h\right)-\cos x}{h}\right)\left[\because \underset{h\to 0}{\lim }\frac{e^x-1}{x}=1\right] \\ =e^{\cos x}\underset{h\to 0}{\lim }\left\{\frac{-2\sin {\displaystyle \left(\frac{x+h+x}{2}\right)}\sin {\displaystyle \left(\frac{x+h-x}{2}\right)}}{h}\right\}\left[\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\right] \\ =e^{\cos x}\underset{h\to 0}{\lim }\frac{-\sin \left({\displaystyle \frac{2x+h}{2}}\right)}{1}\times \frac{\sin \left(\frac{h}{2}\right)}{{\displaystyle \frac{h}{2}}} \\ =e^{\cos x}\underset{h\to 0}{\lim }\frac{-\sin \left({\displaystyle \frac{2x+h}{2}}\right)}{1}\underset{h\to 0}{\times \lim }\frac{\sin \left(\frac{h}{2}\right)}{{\displaystyle \frac{h}{2}}} \\ =e^{\cos x}\underset{h\to 0}{\lim }-\sin \left(\frac{2x+h}{2}\right)\left[\because \frac{\sin x}{x}=1\right] \\ =e^{\cos x}\left(-\sin x\right) \\ =-\sin x{e}^{\cos x} \\ \mathrm{Hence},\frac{d}{dx}\left(e^{\cos x}\right)=-\sin x{e}^{\cos x} \end{gathered}$$