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# Find the derivation of each of the following from the first principle:

Question:

Find the derivation of each of the following from the first principle:

$\tan (3 x+1)$

Solution:

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, $f^{\prime}(x)=3 \sec ^{2}(3 x+1)$