Differentiate

Question:

Differentiate

$\frac{2^{x} \cot x}{\sqrt{x}}$

Solution:

To find: Differentiation of $\left(\frac{2^{x} \cot x}{\sqrt{x}}\right)$

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) $\frac{d \cot x}{d x}=-\operatorname{cosec}^{2} x$

(iii) $\frac{d x^{n}}{d x}=n x^{n-1}$

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

(v) $(u v)^{\prime}=u^{\prime} v+u v^{\prime}$ (Leibnitz or product rule)

Let us take $u=\left(2^{x} \cot x\right)$ and $v=(\sqrt{x})$

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

Applying Product rule

$(g h)^{\prime}=g^{\prime} h+g h^{\prime}$

Taking $g=2^{x}$ and $h=\cot x$

$=\left(2^{x} \log 2\right) \cot x+2^{x}\left(-\operatorname{cosec}^{2} x\right)$

$u^{\prime}=\left(2^{x} \log 2\right) \cot x-2^{x}\left(\operatorname{cosec}^{2} x\right)$

$u^{\prime}=2^{x}\left[\log 2 \cot x-\operatorname{cosec}^{2} x\right]$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\sqrt{x})}{d x}=\frac{1}{2 \sqrt{x}}$

Putting the above obtained values in the formula:-

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

$\left[\frac{2^{x} \cot x}{\sqrt{x}}\right]=\frac{\left\{2^{x}\left[\log 2 \cot x-\operatorname{cosec}^{2} x\right] \times \sqrt{x}\right\}-\left\{\left(2^{x} \cot x\right) \times\left(\frac{1}{2 \sqrt{x}}\right)\right\}}{(\sqrt{x})^{2}}$

$=\frac{\left\{2^{x}\left[\log 2 \cot x-\operatorname{cosec}^{2} x\right] \times \sqrt{x}\right\}-\left\{\left(2^{x} \cot x\right) \times\left(\frac{1}{2 \sqrt{x}}\right)\right\}}{x}$

$=\frac{\left\{2^{x}\left[\log 2 \cot x-\operatorname{cosec}^{2} x\right] \times \sqrt{x}\right\}-\left\{\left(2^{x-1} \cot x\right) \times\left(\frac{1}{\sqrt{x}}\right)\right\}}{x}$

$=\frac{\left\{x 2^{x}\left[\log 2 \cot x-\operatorname{cosec}^{2} x\right]\right\}-\left\{\left(2^{x-1} \cot x\right)\right\}}{x \sqrt{x}}$

$=\frac{\left\{2^{x}\left[x \log 2 \cot x-x \operatorname{cosec}^{2} x\right]\right\}-\left\{\left(2^{x-1} \cot x\right)\right\}}{x^{\frac{3}{2}}}$

Ans $)=\frac{\left\{2^{x}\left[x \log 2 \cot x-x \operatorname{cosec}^{2} x\right]\right\}-\left\{\left(2^{x-1} \cot x\right)\right\}}{x^{\frac{3}{2}}}$