Differentiate

To find: Differentiation of $\frac{2^{x}}{x}$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d a^{x}}{d x}=a^{x} \log a$

Let us take $u=2^{x}$ and $v=x$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(2^{x}\right)}{d x}=2^{x} \log 2$

$v^{\prime}=\frac{d v}{d x}=\frac{d(x)}{d x}=1$

Putting the above obtained values in the formula:

$\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{u}^{\prime} \mathrm{v}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ where $\mathrm{v} \neq 0$ (Quotient rule)

$\left(\frac{2^{x}}{x}\right)^{\prime}=\frac{2^{x} \log 2 \times x-2^{x} \times 1}{(x)^{2}}$

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

Ans $)=\frac{2^{x}(x \log 2-1)}{x^{2}}$