
How do you differentiate $ y=\ln \left( 3{{x}^{2}}-4 \right) $ ?
Answer
463.8k+ views
Hint: We will use the chain rule to differentiate the given expression. We will use the fact that $ \dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)-\dfrac{d}{dx}g\left( x \right) $ . We will also use the derivative of the logarithm function. We know that the derivative of a power function is given by $ \dfrac{d}{dx}\left( {{x}^{n}} \right)=n\cdot {{x}^{n-1}} $ . Using all these formulae and definitions, we will obtain the derivative of the given function.
Complete step by step answer:
The chain rule for differentiating a composition of functions, $ f\circ g\left( x \right) $ , is given by the following
$ {{\left( f\circ g\left( x \right) \right)}^{\prime }}={{f}^{\prime }}\left( g\left( x \right) \right)\cdot {g}'\left( x \right) $
We know that the derivative of the logarithm function is given as $ \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} $ . So, using this and the chain rule, we get the following,
$ \begin{align}
& \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \ln \left( 3{{x}^{2}}-4 \right) \right)\cdot \dfrac{d}{dx}\left( 3{{x}^{2}}-4 \right) \\
& \therefore \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot \dfrac{d}{dx}\left( 3{{x}^{2}}-4 \right) \\
\end{align} $
Now, know that the derivative of difference of functions is the difference in the derivatives of the functions. This means that $ \dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)-\dfrac{d}{dx}g\left( x \right) $ . Using this fact, we can write the above equation as the following,
$ \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot \left( \dfrac{d}{dx}\left( 3{{x}^{2}} \right)-\dfrac{d}{dx}\left( 4 \right) \right) $
We know that the derivative of the constant function is zero. Hence, we have
$ \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot \dfrac{d}{dx}\left( 3{{x}^{2}} \right) $
We know that the derivative of a power function is given by $ \dfrac{d}{dx}\left( {{x}^{n}} \right)=n\cdot {{x}^{n-1}} $ . Using this formula, we get the following,
$ \begin{align}
& \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot 6x \\
& \therefore \dfrac{dy}{dx}=\dfrac{6x}{3{{x}^{2}}-4} \\
\end{align} $
Thus, we have obtained the derivative of the given function.
Note:
It is useful to know the derivatives of standard functions for such type of questions. The chain rule is a very important rule in differentiation. It has important applications when we are dealing with the composition of functions. It is also essential that we understand the differentiation of functions when they are being added, subtracted, etc. These rules are useful in calculations for such type of questions.
Complete step by step answer:
The chain rule for differentiating a composition of functions, $ f\circ g\left( x \right) $ , is given by the following
$ {{\left( f\circ g\left( x \right) \right)}^{\prime }}={{f}^{\prime }}\left( g\left( x \right) \right)\cdot {g}'\left( x \right) $
We know that the derivative of the logarithm function is given as $ \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} $ . So, using this and the chain rule, we get the following,
$ \begin{align}
& \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \ln \left( 3{{x}^{2}}-4 \right) \right)\cdot \dfrac{d}{dx}\left( 3{{x}^{2}}-4 \right) \\
& \therefore \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot \dfrac{d}{dx}\left( 3{{x}^{2}}-4 \right) \\
\end{align} $
Now, know that the derivative of difference of functions is the difference in the derivatives of the functions. This means that $ \dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)-\dfrac{d}{dx}g\left( x \right) $ . Using this fact, we can write the above equation as the following,
$ \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot \left( \dfrac{d}{dx}\left( 3{{x}^{2}} \right)-\dfrac{d}{dx}\left( 4 \right) \right) $
We know that the derivative of the constant function is zero. Hence, we have
$ \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot \dfrac{d}{dx}\left( 3{{x}^{2}} \right) $
We know that the derivative of a power function is given by $ \dfrac{d}{dx}\left( {{x}^{n}} \right)=n\cdot {{x}^{n-1}} $ . Using this formula, we get the following,
$ \begin{align}
& \dfrac{dy}{dx}=\dfrac{1}{3{{x}^{2}}-4}\cdot 6x \\
& \therefore \dfrac{dy}{dx}=\dfrac{6x}{3{{x}^{2}}-4} \\
\end{align} $
Thus, we have obtained the derivative of the given function.
Note:
It is useful to know the derivatives of standard functions for such type of questions. The chain rule is a very important rule in differentiation. It has important applications when we are dealing with the composition of functions. It is also essential that we understand the differentiation of functions when they are being added, subtracted, etc. These rules are useful in calculations for such type of questions.
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