Question

If we have a function as $A = \dfrac{{{2^x}\cot x}}{{\sqrt x }}$, then $\dfrac{{dA}}{{dx}}$ is equal to:
(A) $\dfrac{{{2^x}}}{{\sqrt x }}\left[ {\log 2\cot x - \cos e{c^2}x - \dfrac{{\cot x}}{{2x}}} \right]$
(B) $\dfrac{{{2^x}}}{x}\left[ {\log 2\cot x + \cos e{c^2}x - \dfrac{{\cot x}}{{2x}}} \right]$
(C) $\dfrac{{{2^x}}}{{\sqrt x }}\left[ {\log 2\cot x - \cos e{c^2}x + \dfrac{{\cot x}}{{2x}}} \right]$
(D) None of these

Answer 1

Hint: In the given problem, we are required to differentiate $A = \dfrac{{{2^x}\cot x}}{{\sqrt x }}$ with respect to x. Since, $A = \dfrac{{{2^x}\cot x}}{{\sqrt x }}$ is a rational function in variable x, so we will have to apply quotient rule of differentiation in the process of differentiating the function in x . Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly. We also must know the product rule and quotient rule of differentiation to solve the given problem.

Complete step-by-step solution:
Now, $\left( {\dfrac{{dA}}{{dx}}} \right) = \dfrac{d}{{dx}}\left[ {\dfrac{{{2^x}\cot x}}{{\sqrt x }}} \right]$ .
Now, using the quotient rule of differentiation, we know that $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$ .
So, Applying quotient rule to $\dfrac{d}{{dx}}\left[ {\dfrac{{{2^x}\cot x}}{{\sqrt x }}} \right]$, we get,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{\sqrt x \dfrac{d}{{dx}}\left( {{2^x}\cot x} \right) - \left( {{2^x}\cot x} \right)\dfrac{d}{{dx}}\left( {\sqrt x } \right)}}{{{{\left( {\sqrt x } \right)}^2}}}\]
Substituting the derivative of \[\sqrt x \] with respect to x as $\dfrac{1}{{2\sqrt x }}$,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{\sqrt x \dfrac{d}{{dx}}\left( {{2^x}\cot x} \right) - \left( {{2^x}\cot x} \right) \times \dfrac{1}{{2\sqrt x }}}}{x}\]
Now, applying product rule of differentiation $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) + f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)$, we get,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{\sqrt x \left[ {{2^x} \times \dfrac{d}{{dx}}\left( {\cot x} \right) + \cot x \times \dfrac{d}{{dx}}\left( {{2^x}} \right)} \right] - \left( {{2^x}\cot x} \right) \times \dfrac{1}{{2\sqrt x }}}}{x}\]
Now, we know that the derivative of \[\cot x\] is \[\left( { - \cos e{c^2}x} \right)\]. Also, the derivative of ${2^x}$ with respect to x is ${2^x}\log 2$. So, we get,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{\sqrt x \left[ { - {2^x}\cos e{c^2}x + \cot x \times \left( {{2^x}\log 2} \right)} \right] - \left( {{2^x}\cot x} \right) \times \dfrac{1}{{2\sqrt x }}}}{x}\]
Simplifying the expression, we get,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{ - {2^x}\sqrt x \cos e{c^2}x + \sqrt x \cot x\left( {{2^x}\log 2} \right) - \dfrac{{\left( {{2^x}\cot x} \right)}}{{2\sqrt x }}}}{x}\]
Taking ${2^x}$ common from all the terms,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{{2^x}\left[ { - \sqrt x \cos e{c^2}x + \sqrt x \cot x\log 2 - \dfrac{{\left( {\cot x} \right)}}{{2\sqrt x }}} \right]}}{x}\]
Now, we know that $\sqrt x \times \sqrt x = x$. So, expressing denominator in form of product, we get,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{{2^x}\left[ { - \sqrt x \cos e{c^2}x + \sqrt x \cot x\log 2 - \dfrac{{\left( {\cot x} \right)}}{{2\sqrt x }}} \right]}}{{\sqrt x \times \sqrt x }}\]
Dividing the entire numerator by \[\sqrt x \], we get,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{{2^x}}}{{\sqrt x }}\left[ { - \dfrac{{\sqrt x \cos e{c^2}x}}{{\sqrt x }} + \dfrac{{\sqrt x \cot x\log 2}}{{\sqrt x }} - \dfrac{{\left( {\cot x} \right)}}{{2\sqrt x \times \sqrt x }}} \right]\]
Cancelling common factors in numerator and denominator, we get,
\[ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{{{2^x}}}{{\sqrt x }}\left[ {\log 2\cot x - \cos e{c^2}x - \dfrac{{\cot x}}{{2x}}} \right]\]
Therefore, the derivative of $A = \dfrac{{{2^x}\cot x}}{{\sqrt x }}$ with respect to x is $\dfrac{{{2^x}}}{{\sqrt x }}\left[ {\log 2\cot x - \cos e{c^2}x - \dfrac{{\cot x}}{{2x}}} \right]$.
Hence, option (A) is the correct answer.

Note: The product rule of differentiation involves differentiating a product of two functions and the quotient rule of differentiation involves differentiating a rational function. We must know both of these to solve the given problem. One must know derivatives of some basic functions such as trigonometric and power functions in order to tackle such problems.