Question

Find the derivative of $\dfrac{{{2^x}\cot x}}{{\sqrt x }}$.
A) \dfrac{{{2^x}}}{{\sqrt x }}\left\\{ {\log 2\cot x - {{\csc }^2}x - \dfrac{{\cot x}}{{2x}}} \right\\}
B) \dfrac{{2\left( {x + 1} \right)}}{{\sqrt x }}\left\\{ {\log 2\cot x - {{\csc }^2}x - \dfrac{{\cot x}}{{2x}}} \right\\}
C) \dfrac{{{2^x}}}{{\sqrt x }}\left\\{ {\log 2\cot x - {{\csc }^2}x - \dfrac{{\cot x}}{{{x^2}}}} \right\\}
D) None of these

Answer 1

Here, we will first assume $u = {2^x}$, $v = \cot x$ and $w = {x^{ - \dfrac{1}{2}}}$ in the given equation and then use the product rule for differentiation, $\dfrac{d}{{dx}}\left( {uvw} \right) = u'vw + uv'w + uvw'$ to find the derivate of the given equation.

Complete step by step solution: We are given that the equation is $\dfrac{{{2^x}\cot x}}{{\sqrt x }}$.

We can rewrite the above equation as

$\Rightarrow {2^x}\cot x\left( {{x^{ - \dfrac{1}{2}}}} \right)$

Let us assume that $u = {2^x}$, $v = \cot x$ and $w = {x^{ - \dfrac{1}{2}}}$ in the above equation, we get

We know that the product rule for differentiation is $\dfrac{d}{{dx}}\left( {uvw} \right) = u'vw + uv'w + uvw'$.

First, we will find the first order derivatives of $u$, $v$ and $w$.

$$\Rightarrow u' = \dfrac{d}{{dx}}{2^x} \\\ \Rightarrow u' = {2^x}\log 2 \\\$$ $$\Rightarrow v' = \dfrac{d}{{dx}}\cot x \\\ \Rightarrow v' = - {\csc ^2}x \\\$$ $$\Rightarrow w' = \dfrac{d}{{dx}}{x^{ - \dfrac{1}{2}}} \\\ \Rightarrow w' = - \dfrac{1}{2}{x^{ - \dfrac{1}{2} - 1}} \\\ \Rightarrow w' = - \dfrac{1}{2}{x^{ - \dfrac{3}{2}}} \\\$$

Substituting the above values of $u'$, $v'$ and $w'$ in the above product rule, we get

$$\Rightarrow \dfrac{d}{{dx}}\left( {{2^x}\cot x\left( {{x^{ - \dfrac{1}{2}}}} \right)} \right) = {2^x}\log 2 \cdot \cot x \cdot {x^{ - \dfrac{1}{2}}} + {2^x}\left( { - {{\csc }^2}x} \right){x^{ - \dfrac{1}{2}}} + {2^x}\cot x\left( { - \dfrac{1}{2}{x^{ - \dfrac{3}{2}}}} \right) \\\ \Rightarrow \dfrac{d}{{dx}}\left( {{2^x}\cot x\left( {{x^{ - \dfrac{1}{2}}}} \right)} \right) = {2^x}\log 2 \cdot \cot x \cdot \dfrac{1}{{\sqrt x }} + {2^x}\left( { - {{\csc }^2}x} \right)\dfrac{1}{{\sqrt x }} + {2^x}\cot x\left( { - \dfrac{1}{{2x\sqrt x }}} \right) \\\ \Rightarrow \dfrac{d}{{dx}}\left( {{2^x}\cot x\left( {{x^{ - \dfrac{1}{2}}}} \right)} \right) = \dfrac{{{2^x}}}{{\sqrt x }}\left( {\log 2 \cdot \cot x - {{\csc }^2}x + \dfrac{{\cot x}}{{2x}}} \right) \\\$$

Hence, option A is correct.

Note: Note: While solving these types of problems, students have to rewrite the given equation as ${2^x}\cot x\left( {{x^{ - \dfrac{1}{2}}}} \right)$ and take each expression equal to some variables to use the product rule. Students should know that a derivative of a function of variable$y = f\left( x \right)$ is a measure of the rate at which the value $y$ of the function changes with respect to the change of the variable $x$. The knowledge about basic differentiation rules will be really helpful.