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# If we have a function as $f(x)=\sin (\log x)$ and $y=f\left( \dfrac{2x+3}{3-2x} \right)$, then $\dfrac{dy}{dx}$ is \begin{align} & (A)\sin \left( \log x \right).\dfrac{1}{x\log x} \\ & \left( B \right)\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)\dfrac{12}{9-4{{x}^{2}}} \\ & \left( C \right)\sin \left[ \log \left( \dfrac{2x+3}{3-2x} \right) \right] \\ & \left( D \right)\sin (\log x) \\ \end{align}

Last updated date: 25th Jun 2024
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Hint: To solve this problem, we will need to know the differentiation of composite functions of the form $y=f\left( g\left( x \right) \right)$. The derivative of the composite functions $\dfrac{dy}{dx}$ is evaluated as $y=\dfrac{df\left( g\left( x \right) \right)}{d\left( g\left( x \right) \right)}\times \dfrac{dg\left( x \right)}{dx}$. For this problem, we have the composite function of three different functions. We will use a similar method to differentiate it.

Complete step-by-step solution:
We are given that $f(x)=\sin (\log x)$ and $y=f\left( \dfrac{2x+3}{3-2x} \right)$, we are asked to evaluate $\dfrac{dy}{dx}$. We get the expression for y as, $y=\sin \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)$. As we can see that y is a composite function of the form $f\left( g\left( h\left( x \right) \right) \right)$. Here, $f\left( x \right)=\sin x$, $g\left( x \right)=\log x \And h\left( x \right)=\dfrac{2x+3}{3-2x}$.
As we know that the derivative of $\sin x$ with respect to x is $\cos x$. Thus, the derivatives of $f\left( g\left( h\left( x \right) \right) \right)$ with respect to $g\left( h\left( x \right) \right)$ is $\dfrac{df\left( g\left( h\left( x \right) \right) \right)}{dg\left( h\left( x \right) \right)}=\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)$.
The derivative of $\log x$ with respect to x is $\dfrac{1}{x}$. Thus, the derivative of $g\left( h\left( x \right) \right)$ with respect to $h\left( x \right)$ is $\dfrac{dg\left( h\left( x \right) \right)}{dh\left( x \right)}=\dfrac{1}{\dfrac{2x+3}{3-2x}}=\dfrac{3-2x}{2x+3}$
The derivative of $\dfrac{3-2x}{2x+3}$ with respect to x is $\dfrac{dh(x)}{dx}=\dfrac{2(3-2x)-(-2)(2x+3)}{{{\left( 3-2x \right)}^{2}}}$, simplifying this expression, we get $\dfrac{dh(x)}{dx}=\dfrac{12}{{{\left( 3-2x \right)}^{2}}}$
Using the derivative of a composite function, we can differentiate the function $y=\sin \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)$ as,
$\dfrac{dy}{dx}=\dfrac{df\left( g\left( h\left( x \right) \right) \right)}{dg\left( h\left( x \right) \right)}\times \dfrac{dg\left( h\left( x \right) \right)}{dh\left( x \right)}\times \dfrac{dh(x)}{dx}$
Substituting the values of the derivatives in the above expression, we get
$\Rightarrow \dfrac{dy}{dx}=\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)\times \dfrac{3-2x}{2x+3}\times \dfrac{12}{{{\left( 3-2x \right)}^{2}}}$
Simplifying the above expression, we get
$\Rightarrow \dfrac{dy}{dx}=\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)\dfrac{12}{\left( 3-2x \right)\left( 2x+3 \right)}$
$\Rightarrow \dfrac{dy}{dx}=\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)\dfrac{12}{9-4{{x}^{2}}}$
Thus, the derivative of the given expression is $\dfrac{dy}{dx}=\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)\dfrac{12}{9-4{{x}^{2}}}$.
Hence, the answer is an option $\left( B \right)$.

Note: We can use this method to find the derivative of a composite function having more functions than this. In general words, $y=f\left( g(h(......(x))) \right)$ can be differentiated by differentiating the function outside with respect to the function inside it.