Question
Differentiate the following functions from first principles:
$\text{e}^{\cos\text{x}}$
$\text{e}^{\cos\text{x}}$
✓
Answer
Let $\text{f(x)}=\text{e}^{\cos\text{x}}$
$\Rightarrow\text{f}(\text{x}+\text{h})=\text{e}^{\cos(\text{x}+\text{h})}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\cos(\text{x}+\text{h})}-\text{e}^{\cos\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\cos(\text{x}+\text{h})-\cos\text{x}}\Big]\times\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\times\Big(\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}\Big)$
$\Big[\text{Since},\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\ \text{e}^{\cos\text{x}}\times\bigg(\frac{-2\sin\frac{\text{x}+\text{h}+\text{x}}{2}\times\sin\frac{\text{x}+\text{h}-\text{x}}{2}}{\text{h}}\bigg)$
$\begin{bmatrix} \cos\text{A}-\cos\text{B}=-2\sin\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2} \end{bmatrix}$
$=\text{e}^{\cos\text{x}}\lim\limits_{\text{h}\rightarrow0}\frac{-\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{2}\times\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\frac{\text{h}}{2}}$
$=\text{e}^{\cos\text{x}}\lim\limits-2\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)\times\frac{1}{2}$
$\Big[\text{Since},\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\text{e}^{\cos\text{x}}(-\sin\text{x})$
$=-\sin\text{xe}^{\cos\text{x}}$
Hence,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}}\big)=-\sin\text{xe}^{\cos\text{x}}$
$\Rightarrow\text{f}(\text{x}+\text{h})=\text{e}^{\cos(\text{x}+\text{h})}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\cos(\text{x}+\text{h})}-\text{e}^{\cos\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\Big[\frac{\text{e}^{\cos(\text{x}+\text{h})-\cos\text{x}}-1}{\cos(\text{x}+\text{h})-\cos\text{x}}\Big]\times\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{ e}^{\cos\text{x}}\times\Big(\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}\Big)$
$\Big[\text{Since},\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\ \text{e}^{\cos\text{x}}\times\bigg(\frac{-2\sin\frac{\text{x}+\text{h}+\text{x}}{2}\times\sin\frac{\text{x}+\text{h}-\text{x}}{2}}{\text{h}}\bigg)$
$\begin{bmatrix} \cos\text{A}-\cos\text{B}=-2\sin\frac{\text{A}+\text{B}}{2}\sin\frac{\text{A}-\text{B}}{2} \end{bmatrix}$
$=\text{e}^{\cos\text{x}}\lim\limits_{\text{h}\rightarrow0}\frac{-\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)}{2}\times\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\frac{\text{h}}{2}}$
$=\text{e}^{\cos\text{x}}\lim\limits-2\sin\Big(\frac{2\text{x}+\text{h}}{2}\Big)\times\frac{1}{2}$
$\Big[\text{Since},\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\text{e}^{\cos\text{x}}(-\sin\text{x})$
$=-\sin\text{xe}^{\cos\text{x}}$
Hence,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}}\big)=-\sin\text{xe}^{\cos\text{x}}$
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