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

Hint: 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.