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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
\[\begin{align}
  & (A)\sin \left( \log x \right).\dfrac{1}{x\log x} \\
 & \left( B \right)\cos \left( \log \left( \dfrac{2x+3}{3-2x} \right) \right)\dfrac{12}{9-4{{x}^{2}}} \\
 & \left( C \right)\sin \left[ \log \left( \dfrac{2x+3}{3-2x} \right) \right] \\
 & \left( D \right)\sin (\log x) \\
\end{align}\]

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Last updated date: 25th Jun 2024
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Answer
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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.