Question

Find the derivative of $\tan \left(ax+b\right)$ from the first principle?

Answer 1

Hint: There are two ways to do this question. One way is by directly applying the formula or we can say chain rule of differentiation. The other way is doing it by using the first principle. In first principle we have to use a formula $$f^{\prime }(x)=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$$ where $f(x)=\tan (ax+b)$ to do this question.

Formula used: $$f^{\prime }(x)=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$$

Complete step by step solution:
Using the derivative definition, if:
$$\Rightarrow f(x)=tan(ax+b)$$
Then, the derivative f'(x) is given by:
$$f^{\prime }(x)=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$$
$$\Rightarrow li{m}_{h\to 0}={\displaystyle \frac{tan(a(x+h)+b)-tan(ax+b)}{h}}$$
Now using the formula $\tan \left(x+y\right)={\displaystyle \frac{\tan x+\tan y}{1-\tan x\tan y}}$
$$\Rightarrow li{m}_{h\to 0}={\displaystyle \frac{{\displaystyle \frac{tan(ax+b)+tanah}{1-tan(ax+b)tanah}}-tan(ax+b)}{h}}$$
Now, taking L.C.M
$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{{\displaystyle \frac{tan(ax+b)+tanah-tan\left(ax+b\right)(1-tan(ax+b)tanah)}{1-tan(ax+b)tanah}}}{h}}$
On simplification, we get
$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{tan(ax+b)+tanah-tan(ax+b)+tan2(ax+b)tanah}{h(1-tan(ax+b)tanah)}}$
$$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{tanah(1+ta{n}^2(ax+b))}{h(1-tan(ax+b)tanah)}}$$
Now, separating the terms
$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{1+ta{n}^2(ax+b)}{1-tan(ax+b)tanah}}\cdot {\displaystyle \frac{tanah}{h}}$
$$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{1+tan2(ax+b)}{1-tan(ax+b)tanah}}\cdot li{m}_{h\to 0}{\displaystyle \frac{tanah}{h}}$$
Consider the first limit
$$L_1=li{m}_{h\to 0}{\displaystyle \frac{1+tan2(ax+b)}{1-tan(ax+b)tanah}}$$
$$\Rightarrow {\displaystyle \frac{1+tan2(ax+b)}{1-tan(ax+b)0}}$$
$$\Rightarrow 1+ta{n}^2(ax+b)$$
$$\Rightarrow se{c}^2(ax+b)$$
And, now the second limit:
$$L_2=li{m}_{h\to 0}{\displaystyle \frac{tanah}{h}}$$
$$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{sinah}{cosah}}\cdot {\displaystyle \frac{1}{h}}$$
$$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{sinah}{h}}\cdot {\displaystyle \frac{1}{cosah}}$$
$$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{asinah}{ah}}\cdot {\displaystyle \frac{1}{cosah}}$$
Now, splitting the limits
$$\Rightarrow li{m}_{h\to 0}{\displaystyle \frac{asinah}{ah}}\cdot li{m}_{h\to 0}{\displaystyle \frac{1}{cosah}}$$
$$\Rightarrow ali{m}_{\theta \to 0}{\displaystyle \frac{sin\theta }{\theta }}\cdot li{m}_{h\to 0}{\displaystyle \frac{1}{cosah}}$$
And for this limit we have:
$$\Rightarrow li{m}_{\theta \to 0}{\displaystyle \frac{sin\theta }{\theta }}=1andli{m}_{h\to 0}{\displaystyle \frac{1}{cosah}}=1$$
Leading to:
$$L_2=a$$
Combining these results, we have
$\Rightarrow f^{\prime }(x)=se{c}^2(ax+b)\cdot a$
$\Rightarrow ase{c}^2(ax+b)$
Therefore, the required derivative is $ase{c}^2(ax+b)$.

Note:
Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to $$f^{\prime }(x)=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$$