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

Provirde solution RD Sharma maths class 12 chapter 10 differentiationability  exercise 10 .1 question 1 maths textbook solution

Answers (1)

Answer:

\left ( -e^{-x} \right )

Hint:

Use first principle to find the differentiation of 

\left ( -e^{-x} \right )

Given:

\left ( -e^{-x} \right )

 

Solution:

Let

f\left ( x \right )= e^{-x}

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

So, we will use formula of differentiation by first principle,

\therefore \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}= \lim_{h\rightarrow o}\frac{e^{-\left ( x+h \right )}-e^{-x}}{h}

                            =\lim_{h\rightarrow 0}\frac{e^{-x}\times e^{-h}-e^{-x}}{h}  

                            =\lim_{h\rightarrow 0}\frac{e^{-x}\left ( e^{-h}-1 \right )}{h}

                             =\lim_{h\rightarrow 0}\frac{e^{-x}\left ( e^{-h}-1 \right )}{-h}\times \left ( -1 \right )

                              = \lim_{h\rightarrow 0}e^{-x}\times \left ( \frac{e^{-h}-1}{-h} \right )\times \left ( -1 \right )

                            

                            = \lim_{h\rightarrow 0}e^{-x}\left ( 1 \right )\times \left ( -1 \right )                                                 \left [ \therefore \lim_{h\rightarrow 0}\frac{e^{x}-1}{x} \right ]= 1

                           =\left ( -e^{-x} \right )

Hence, the differentiation of e^{-x} is \left ( -e^{-x} \right )

 

 

 

 

 

                       




 

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