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  • Please,please help me - Limit , continuity and differentiability - JEE Main-10

Let y = (x + 1)(x + 2)(x + 3)(x + 4)(x + 5) then \frac{\mathrm{d} y}{\mathrm{d} x} at x =0 equals ?

  • Option 1)

    274

  • Option 2)

    234

  • Option 3)

    224

  • Option 4)

    244

 

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)+....]

 

 

Takeing log both sides

log_e y = log_e (x+1) + log_e (x+2) + log_e (x+3) + log_e (x+4) + log_e (x+5)

Now differenciating both sides,

\frac{1}{y}\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + \frac{1}{x+4} + \frac{1}{x+5}

\left [\frac{\mathrm{d} y}{\mathrm{d} x} \right ]_{(x = 0)} = y(0) \left \{ \frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{5}\right \} = 120 + 60 + 30 + 24 = 274


Option 1)

274

Option 2)

234

Option 3)

224

Option 4)

244

Posted by

Himanshu

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