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Provide solution RD Sharma maths class 12 chapter 10 differentiability exercise 10.1 question 4 maths textbook solution

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-\left ( \sin x \right )\cdot e^{\cos x}


Use first principle to find the differentiation


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



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

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

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

Take e^{\cos x} common from numerator

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

                       =\lim_{h\rightarrow 0}\frac{e^{\cos x}\left [ e^{\cos x\left ( x+h \right )-\cos x} -1\right ]}{h}                                              \left [ \because \frac{a^{m}}{a^{n}}=a^{m-n} \right ]                        


Multiply and divide \left [ \cos \left ( x+h \right )-\cos x \right ]

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


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


                 =\lim_{h\rightarrow 0}e^{\cos x}\times \left ( \frac{\cos \left ( x+h \right )-\cos x}{h} \right )                \left [ \because \cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2} \right ]                      

                =\lim_{h\rightarrow 0}e^{\cos x}\times \left [ \frac{-2\sin \frac{x+h+x}{2}\times \sin \frac{x+h-x}{2}}{h} \right ]                

e^{\cos x} Is a function of f\left ( x \right ) and limit is applied on variable ‘h’

So,e^{\cos x} can be taken outside

= e^{\cos x}\times \lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{2x+h}{2}\ \right )\times \sin \left ( \frac{h}{2} \right )}{h}

Multiply the denominator by 2 and divide the denominator by 2

= e^{\cos x}\times \lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{2x+h}{2} \right )}{2}\times \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}

= e^{\cos x}\times \lim_{h\rightarrow 0}\left [ -\sin \left ( \frac{2x+h}{2} \right ) \right ]\times 1                                        \left [ \because \lim_{h\rightarrow 0}\frac{\sin x}{x}=1 \right ]

=e^{\cos x}\times\left ( -\sin x \right )=-\sin xe^{\cos x}

Hence, the differentiation of e^{\cos x} is \left [ -\sin xe^{\cos x} \right ]

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