Find the derivation of each of the following from the first principle:
$\tan (3 x+1)$
Let f(x) = tan (3x + 1)
We need to find the derivative of f(x) i.e. f’(x)
We know that,
$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{\mathrm{h}}$ …(i)
$f(x)=\tan (3 x+1)$
$f(x+h)=\tan [3(x+h)+1]$
Putting values in (i), we get
$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\tan [3(\mathrm{x}+\mathrm{h})+1]-\tan [3 \mathrm{x}+1]}{\mathrm{h}}$
Using the formula:
$\tan A-\tan B=\frac{\sin (A-B)}{\cos A \cos B}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\frac{\sin [3(\mathrm{x}+\mathrm{h})+1-(3 \mathrm{x}+1)]}{\cos [3(\mathrm{x}+\mathrm{h})+1] \cos [3 \mathrm{x}+1]}}{\mathrm{h}}$
$=\lim _{h \rightarrow 0} \frac{\frac{\sin [3 x+3 h+1-3 x-1]}{\cos [3(x+h)+1] \cos [3 x+1]}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin 3 h}{h[\cos [3(x+h)+1] \cos [3 x+1]]}$
$=\lim _{h \rightarrow 0} \frac{\sin 3 h}{h} \times \lim _{h \rightarrow 0} \frac{1}{\cos [3(x+h)+1] \cos [3 x+1]}$
$=\lim _{h \rightarrow 0} \frac{\sin 3 h}{3 h} \times 3 \times \lim _{h \rightarrow 0} \frac{1}{\cos [3(x+h)+1] \cos [3 x+1]}$
$=3(1) \times \lim _{h \rightarrow 0} \frac{1}{\cos [3(x+h)+1] \cos [3 x+1]}$
$\left[\because \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}=1\right]$
Putting h = 0, we get
$=3 \times \frac{1}{\cos [3(x+0)+1] \cos [3 x+1]}$
$=\frac{3}{\cos [3 x+1] \cos [3 x+1]}$
$=\frac{3}{\cos ^{2}(3 x+1)}$
$=3 \sec ^{2}(3 x+1)\left[\because \frac{1}{\cos x}=\sec x\right]$
Hence, $f^{\prime}(x)=3 \sec ^{2}(3 x+1)$