The correct option is C We konw,d/dx(x.2^x 1) d/dx(x.2^x) d/dx(1) We know, d/dx(u.v)=u d/dx v+v d/d ...

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Logarithmic Inequalities

d/d xx .2x-1=

Question

$\frac{d}{d x} (x {.2}^{x} - 1) =$

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Solution

The correct option is C

We konw, $\frac{d}{d x} (x {.2}^{x} - 1)$

$\frac{d}{d x} (x {.2}^{x}) - \frac{d}{d x} (1)$

We know, $\frac{d}{d x} (u . v) = u \frac{d}{d x} v + v \frac{d}{d x} u$

$\Rightarrow x \frac{d}{d x} 2^{x} + 2^{x} . \frac{d}{d x} x - 0$

$= x {.2}^{x} . l o g_{e}^{2} + 2^{x}$ ${\frac{d}{d x} a^{x} = a^{x} l o g_{e}^{a}}$

$= 2^{x} (x . l o g_{e}^{2} + 1)$

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