Question

If $y = b\cos \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)$, then $\dfrac{{dy}}{{dx}}$equals
A. $ - nb\sin \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)$
B. $nb\sin \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)$
C. $ - \dfrac{{nb}}{x}\sin \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)$
D. None of these

Answer 1

Hint: In the given problem, we are required to differentiate $y = b\cos \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)$ with respect to x. The given function is a composite function, so we will have to apply chain rule of differentiation in the process of differentiation. So, differentiation of $y = b\cos \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)$ with respect to x will be done layer by layer. Also, we will use properties of logarithmic functions to simplify the differentiation of the function.

Formula used:
1. $\log {p^q} = q\log p$
2. Chain rule: $\dfrac{d}{dx} f(g(x)) = f^{‘}(g(x)).g^{‘}(x)$
3. $\dfrac{d}{dx} \cos x = - sin x$
4. $\dfrac{d}{dx} \log x = \dfrac{1}{x}$

Complete step by step solution:
We are given, $y = b\cos \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)$.
So, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {b\cos \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)} \right]$
Now, we use the property of logarithm $\log {p^q} = q\log p$.
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {b\cos \left( {n\log \left( {\dfrac{x}{n}} \right)} \right)} \right]$
Taking constant b out of differentiation, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = b\dfrac{d}{{dx}}\left[ {\cos \left( {n\log \left( {\dfrac{x}{n}} \right)} \right)} \right]$
Now, Let us assume $u = \left( {n\log \left( {\dfrac{x}{n}} \right)} \right)$. So substituting $\left( {n\log \left( {\dfrac{x}{n}} \right)} \right)$ as $u$, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = b\dfrac{d}{{dx}}\left[ {\cos \left( u \right)} \right] = b\dfrac{d}{{du}}\left[ {\cos \left( u \right)} \right] \times \dfrac{{du}}{{dx}}$
Derivative of $\cos u$ with respect to u is $ - \sin u$. So, we have,
$ \Rightarrow \dfrac{{dy}}{{dx}} = b\left[ { - \sin \left( u \right)} \right] \times \dfrac{{du}}{{dx}}$
Now, putting back $u$ as $\left( {n\log \left( {\dfrac{x}{n}} \right)} \right)$, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = b\left[ { - \sin \left( {\left( {n\log \left( {\dfrac{x}{n}} \right)} \right)} \right)} \right] \times \dfrac{d}{{dx}}\left( {n\log \left( {\dfrac{x}{n}} \right)} \right)\] because \[\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {n\log \left( {\dfrac{x}{n}} \right)} \right)\]
Taking constants out of the differentiation, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - b\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right) \times n\dfrac{d}{{dx}}\left( {\log \left( {\dfrac{x}{n}} \right)} \right)\]
Now, Let us assume $v = \left( {\dfrac{x}{n}} \right)$. So, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - b\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right) \times n\dfrac{d}{{dx}}\left( {\log v} \right)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - b\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right) \times n\dfrac{d}{{dv}}\left( {\log v} \right) \times \dfrac{{dv}}{{dx}}\]
Now, we know that derivative of $\log y$ with respect to y is \[\dfrac{1}{y}\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - b\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right) \times \dfrac{n}{v} \times \dfrac{{dv}}{{dx}}\]
Now, putting back $v = \left( {\dfrac{x}{n}} \right)$, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - b\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right) \times \dfrac{n}{{\left( {\dfrac{x}{n}} \right)}} \times \dfrac{d}{{dx}}\left( {\dfrac{x}{n}} \right)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - b\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right) \times \dfrac{n}{{\left( {\dfrac{x}{n}} \right)}} \times \dfrac{d}{{dx}}\left( {\dfrac{x}{n}} \right)\]
So, substituting the equivalent expression of $\dfrac{d}{{dx}}\left( {\dfrac{x}{n}} \right)$, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - b\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right) \times \dfrac{{{n^2}}}{x} \times \dfrac{1}{n}\]
Simplifying the product of expression, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{bn}}{x}\sin \left( {n\log \left( {\dfrac{x}{n}} \right)} \right)\]
Again, using the logarithmic property $\log {p^q} = q\log p$, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{bn}}{x}\sin \left( {\log {{\left( {\dfrac{x}{n}} \right)}^n}} \right)\]

So, option (C) is the correct answer.

Note: The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using the chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.