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  • Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10 point 1 question 1 maths textbook solution

Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 1 maths textbook solution

Answers (1)

Answer:

ae^{\left ( ax+b \right )}

Hint:

Use first principle to find the differentiation

Given:

\left ( e^{ax+b} \right )

Solution:

Let

 

f\left ( x \right )=e^{ax+b}

f\left ( x+h \right )=e^{a\left ( x+h \right )+b}

Now, we will use formula of differentiation by first principle

\frac{d}{dx}\left ( f\left ( x \right ) \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}

\frac{d}{dx}\left ( e^{ax+b} \right )=\lim_{h\rightarrow 0}\frac{e^{a\left ( x+h \right )+b}-e^{ax+b}}{h}

=\lim_{h\rightarrow 0}\frac{e^{ax+ah+b}-e^{ax+b}}{h}

=\lim_{h\rightarrow 0}\frac{\left ( e^{ax+b} \right )\left ( e^{ah} \right )-e^{ax+b}}{h}

=\lim_{h\rightarrow 0}\frac{e^{ax+b}\left ( e^{ah}-1 \right )}{h}

Multiply and divide by a

=\lim_{h\rightarrow 0}e^{ax+b}\frac{\left ( e^{ah}-1 \right )}{ah}\times a

=\lim_{h\rightarrow 0}\left ( e^{ax+b} \right )\times a                                        \left [ \because \lim_{h\rightarrow 0}\frac{e^{x}-1}{x}=1\right ]         

=ae^{ax+b}

Hence, the differentiation of e^{ax+b} is ae^{ax+b}

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