Sets1

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  • #66965
    Shreya Nair
    Participant

    Sets1

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    #71214
    venkat jothi
    Participant

    given that f(x) is a differentiable function.

    $$lim_{\delta \to 0} \big(\frac{f(x+\delta x)}{f(x)}\big)^{\frac{1}{\delta}} ———(1)$$

    by direct substitution it is in the form of $1^\infty$.

    so we can use this formula

    $$lim_{\delta \to 0} \big(1+g(\delta)\big)^{h(\delta)}=e^{lim_{\delta \to 0} g(\delta).h(\delta)$$

    here , from(1) we get

    $$\displaystyle \Rightarrow lim_{\delta \to 0} \big(1+\frac{f(x+\delta x)}{f(x)} -1\big)^{\frac{1}{\delta}}$$

    $$\displaystyle \Rightarrow lim_{\delta \to 0} \big(1+\frac{f(x+\delta x)-f(x)}{f(x)}\big)^{\frac{1}{\delta}}$$

    here, $g(\delta)=\frac{f(x+\delta x)-f(x)}{f(x)}$ and $h(\delta )= \frac{1}{\delta}$

    $$\displaystyle \Rightarrow e^{lim_{\delta \to 0} g(\delta).h(\delta)$$

    $$\displaystyle \Rightarrow e^{lim_{\delta \to 0}\frac{f(x+\delta x)-f(x)}{f(x)} .\frac{1}{\delta}$$

    by rearranging terms

    $$\Rightarrow e^{lim_{\delta \to 0}\frac{f(x+\delta x)-f(x)}{\delta} .\frac{1}{f(x)}$$

    by replacing the Definition of derivative $lim_{\delta \to 0}\frac{f(x+\delta x)-f(x)}{\delta} =f'(x)$. we get

    $$\Rightarrow e^{\frac{f'(x)}{f(x)}}$$

     

    • This reply was modified 1 week, 2 days ago by venkat jothi.
    #71261
    venkat jothi
    Participant

    SORRY LaTeX  DOES NOT WORK  FOR SOME EQUATIONS IN MY ABOVE SOLUTION

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