CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 83 Marks
Question
Differentiate $e ^{ ax + b }$ from first principle.
✓
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
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a(z+h)+b}-e^{a z+b}}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a z+b} e^{a k}-e^{a z+b}}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{a z+b}\left(e^{a h}-1\right)}{h}$
$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} e^{a x+b} \times \lim _{h \rightarrow 0} \frac{e^{a h}-1}{h}$
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: $\lim _{x \rightarrow 0} \frac{e^2-1}{x}=\log _e e =1$
$\Rightarrow f^{\prime}(x)=e^{a x+b} \times(a)$
$\Rightarrow f^{\prime}(x)=a e^{a x+b}$
Hence,
Derivative of $f(x)=e^{a x+b}=a e^{a x+b}$
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