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  • Help me please, - Limit , continuity and differentiability - JEE Main-8

Let y = \frac{(x+1)^{2}(x+2)}{(x+3)}then \frac{\mathrm{d} y }{\mathrm{d} x}, at x = 0 equals ?

  • Option 1)

    \frac{9}{}4

  • Option 2)

    \frac{4}{}9

  • Option 3)

    \frac{9}{}{13}

  • Option 4)

    \frac{13}{}9

 

Answers (1)

best_answer

As we have learnt,

 

Logarithmic differentiation -

When a function consists of the product of the quotient of a number  of functions we take logarithm.such as

1.\:\:\: y={f_{1}(x)}^{f_{2}(x)}
 

2.\:\:\: y=f_{1}(x).f_{2}(x).f_{3}(x)......
 

2.\:\:\: y=\frac{f_{1}(x)\:f_{2}(x)\:f_{3}(x)....}{\phi_{1} (x)\:\phi_{2} (x)\:\phi _{3}(x)....}

- wherein

For

1.\:\:log\:y=f_{2}(x)log\:f_{1}(x)
 

2.\:\:log\:y=log\:f_{1}(x)+log\:f_{2}(x)+log\:f_{3}(x)+......
 

2.\:\:log\:y=[log\:f_{1}(x)+log\:f_{2}(x)+.....]-[log\:\phi _{1}(x)+log\:\phi _{2}(x)+....]

 

 Taking log on both sides and then differenciating,

\log_{e} y = 2log_e (x + 1) + log_e(x +2) - log_e(x + 3) \\*\Rightarrow \frac{1}{y} \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2}{x + 1} + \frac{1}{x +2}- \frac{1}{x + 3} \\*\Rightarrow \left[\frac{\mathrm{d} y}{\mathrm{d} x} \right ]_{(x = 0)} = y(0)\left (\frac{2}{1} + \frac{1}{2}- \frac{1}{3} \right ) = \frac{26}{18} = \frac{13}{9}

 


Option 1)

\frac{9}{}4

Option 2)

\frac{4}{}9

Option 3)

\frac{9}{}{13}

Option 4)

\frac{13}{}9

Posted by

Himanshu

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