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Question: Find the derivative of \(\tan \left( {ax + b} \right)\) from the first principle?...

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

Explanation

Solution

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′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = {\lim _{h \to 0}}\dfrac{{f(x + h) - f(x)}}{h} where f(x)=tan⁡(ax+b)f(x) = \tan (ax + b) to do this question.

Formula used: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = {\lim _{h \to 0}}\dfrac{{f(x + h) - f(x)}}{h}

Complete step by step solution:
Using the derivative definition, if:
⇒f(x)=tan(ax+b)\Rightarrow f(x) = tan(ax + b)
Then, the derivative f'(x) is given by:
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = {\lim _{h \to 0}}\dfrac{{f(x + h) - f(x)}}{h}
⇒limh→0=tan(a(x+h)+b)−tan(ax+b)h\Rightarrow li{m_{h \to 0}} = \dfrac{{tan(a(x + h) + b) - tan(ax + b)}}{h}
Now using the formula tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}
⇒limh→0=tan(ax+b)+tanah1−tan(ax+b)tanah−tan(ax+b)h\Rightarrow li{m_{h \to 0}} = \dfrac{{\dfrac{{tan(ax + b) + tanah}}{{1 - tan(ax + b)tanah}} - tan(ax + b)}}{h}
Now, taking L.C.M
⇒limh→0tan(ax+b)+tanah−tan(ax+b)(1−tan(ax+b)tanah)1−tan(ax+b)tanahh\Rightarrow li{m_{h \to 0}}\dfrac{{\dfrac{{tan(ax + b) + tanah - tan\left( {ax + b} \right)(1 - tan(ax + b)tanah)}}{{1 - tan(ax + b)tanah}}}}{h}
On simplification, we get
⇒limh→0tan(ax+b)+tanah−tan(ax+b)+tan2(ax+b)tanahh(1−tan(ax+b)tanah)\Rightarrow li{m_{h \to 0}}\dfrac{{tan(ax + b) + tanah - tan(ax + b) + tan2(ax + b)tanah}}{{h(1 - tan(ax + b)tanah)}}
⇒limh→0tanah(1+tan2(ax+b))h(1−tan(ax+b)tanah)\Rightarrow li{m_{h \to 0}}\dfrac{{tanah(1 + ta{n^2}(ax + b))}}{{h(1 - tan(ax + b)tanah)}}
Now, separating the terms
⇒limh→01+tan2(ax+b)1−tan(ax+b)tanah⋅tanahh\Rightarrow li{m_{h \to 0}}\dfrac{{1 + ta{n^2}(ax + b)}}{{1 - tan(ax + b)tanah}} \cdot \dfrac{{tanah}}{h}
⇒limh→01+tan2(ax+b)1−tan(ax+b)tanah⋅limh→0tanahh\Rightarrow li{m_{h \to 0}}\dfrac{{1 + tan2(ax + b)}}{{1 - tan(ax + b)tanah}} \cdot li{m_{h \to 0}}\dfrac{{tanah}}{h}
Consider the first limit
L1=limh→01+tan2(ax+b)1−tan(ax+b)tanah{L_1} = li{m_{h \to 0}}\dfrac{{1 + tan2(ax + b)}}{{1 - tan(ax + b)tanah}}
⇒1+tan2(ax+b)1−tan(ax+b)0\Rightarrow \dfrac{{1 + tan2(ax + b)}}{{1 - tan(ax + b)0}}
⇒1+tan2(ax+b)\Rightarrow 1 + ta{n^2}(ax + b)
⇒sec2(ax+b)\Rightarrow se{c^2}(ax + b)
And, now the second limit:
L2=limh→0tanahh{L_2} = li{m_{h \to 0}}\dfrac{{tanah}}{h}
⇒limh→0sinahcosah⋅1h\Rightarrow li{m_{h \to 0}}\dfrac{{sinah}}{{cosah}} \cdot \dfrac{1}{h}
⇒limh→0sinahh⋅1cosah\Rightarrow li{m_{h \to 0}}\dfrac{{sinah}}{h} \cdot \dfrac{1}{{cosah}}
⇒limh→0asinahah⋅1cosah\Rightarrow li{m_{h \to 0}}\dfrac{{asinah}}{{ah}} \cdot \dfrac{1}{{cosah}}
Now, splitting the limits
⇒limh→0asinahah⋅limh→01cosah\Rightarrow li{m_{h \to 0}}\dfrac{{asinah}}{{ah}} \cdot li{m_{h \to 0}}\dfrac{1}{{cosah}}
⇒a  limθ→0sinθθ⋅limh→01cosah\Rightarrow a\;li{m_{\theta \to 0}}\dfrac{{sin\theta }}{\theta } \cdot li{m_{h \to 0}}\dfrac{1}{{cosah}}
And for this limit we have:
⇒limθ→0sinθθ=1  and  limh→01cosah=1\Rightarrow li{m_{\theta \to 0}}\dfrac{{sin\theta }}{\theta } = 1\;and\;li{m_{h \to 0}}\dfrac{1}{{cosah}} = 1
Leading to:
L2=a{L_2} = a
Combining these results, we have
⇒f′(x)=sec2(ax+b)⋅a\Rightarrow f'(x) = se{c^2}(ax + b) \cdot a
⇒asec2(ax+b)\Rightarrow ase{c^2}(ax + b)
Therefore, the required derivative is asec2(ax+b)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′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = {\lim _{h \to 0}}\dfrac{{f(x + h) - f(x)}}{h}