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Solve! - Limit , continuity and differentiability - JEE Main-14

Let f(x)= x^{3} log _{e} x   then f'(x) equals 

  • Option 1)

    x^{2}(1+3log _{e}x)

  • Option 2)

    x^{2}

  • Option 3)

    3x^{2}log _{e}x

  • Option 4)

    x^{2}+log _{e}x

 
Answers (1)
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As we have learned

Product Rule for differentiation -

\frac{d}{dx}{f(x).g(x)}=f(x).\frac{d}{dx}g(x)+g(x).\frac{d}{dx}f(x)

 

\Rightarrow \frac{d}{dx}(u.v)=u.\frac{dv}{dx}+v.\frac{du}{dx}

- wherein

Take only one function for derivative along with other function.

 

 f'(x)= \frac{d}{dx}(x^{3})log_{e}x= x^{3} \frac{d}{dx}(log_{e}x)+ log _{e}x \frac{d}{dx} (x^{3})

f'(x)=x^{3}\cdot \frac{1}{x}+(log_{e}x)\cdot (3x^{2})

\Rightarrow f'(x)=x^{2}(1+3log_{e}x)

 

 

 

 

 


Option 1)

x^{2}(1+3log _{e}x)

Option 2)

x^{2}

Option 3)

3x^{2}log _{e}x

Option 4)

x^{2}+log _{e}x

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