Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 83 Marks
Question
Differentiate $e ^{ ax + b }$ from first principle.
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Answer
We need to find derivative of $f(x)=e^{a x+b}$ Derivative of a function f(x) is given by $f ^{\prime}( x )=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\{$ where h is a very small positive number $\}$ $\therefore$ derivative of $f(x)=e^{a x+b}$ is given as $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $\begin{array}{l}\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a(z+h)+b}-e^{a z+b}}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a z+b} e^{a k}-e^{a z+b}}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a z+b}\left(e^{a h}-1\right)}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} e^{a x+b} \times \lim _{h \rightarrow 0} \frac{e^{a h}-1}{h}\end{array}$ As one of the limits $\times \lim _{h \rightarrow 0} \frac{e^{h h}-1}{h}$ can't be evaluated by directly putting the value of $h$ as it will take $\frac{0}{0}$ form. So we need to take steps to find its value. $\Rightarrow f ^{\prime}( x )=\lim _{ h \rightarrow 0} e ^{ ax + b } \times \lim _{h \rightarrow 0} \frac{e^{h h}-1}{a h} \times a$ Use the formula: $\lim _{x \rightarrow 0} \frac{e^2-1}{x}=\log _e e =1$ $\begin{array}{l}\Rightarrow f^{\prime}(x)=e^{a x+b} \times(a) \\ \Rightarrow f^{\prime}(x)=a e^{a x+b}\end{array}$ Hence, Derivative of $f(x)=e^{a x+b}=a e^{a x+b}$
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