Question

How do you find the derivative of $\ln \left( {{x}^{2}}-4 \right)$ ?

Answer 1

To find the derivative of $\ln \left( {{x}^{2}}-4 \right)$, we are going to use the chain rule. In which we will first take the derivative of $\ln \left( {{x}^{2}}-4 \right)$ with respect to x by assuming ${{x}^{2}}-4$ as x. Then the derivative will be the same as that of $\ln x$. After that, we are going to find the derivative of ${{x}^{2}}-4$ with respect to x and multiply the result of this derivative with the previous one.

Complete answer:
The function in x which we are asked to take the derivative of is:
$\ln \left( {{x}^{2}}-4 \right)$
Now, we are going to derive the above function by assuming ${{x}^{2}}-4$ as x then the derivative of the above function is the same as that of $\ln x$ with respect to x.
$\dfrac{1}{{{x}^{2}}-4}$ ……………………...… (1)
We know that derivative of $\ln x$ with respect to x is equal to $\dfrac{1}{x}$ and the x in the above function is ${{x}^{2}}-4$ so in the above we have written ${{x}^{2}}-4$ in place of x.
Now, we are going to take the derivative of ${{x}^{2}}-4$ which is equal to:
$2x$ …………. (2)
And now, multiplying (1) and (2) we get,
$\implies\dfrac{1}{{{x}^{2}}-4}\left( 2x \right)$
Rearranging the above expression we get,
$\implies\dfrac{2x}{{{x}^{2}}-4}$
Hence, from the above, we have calculated the derivative of $\ln \left( {{x}^{2}}-4 \right)$ as $\dfrac{2x}{{{x}^{2}}-4}$.

Note: The other way of solving the above problem is as follows:
Let us assume ${{x}^{2}}-4=t$ in the above function. Then the derivative of $\ln \left( {{x}^{2}}-4 \right)$ will look like:
$\dfrac{d\ln \left( {{x}^{2}}-4 \right)}{dx}$
Converting the above derivative in terms of t by multiplying and dividing by $dt$ we get,
$\dfrac{d\ln \left( {{x}^{2}}-4 \right)}{dx}\times \dfrac{dt}{dt}$
Now, rearranging the above expression we get,
$\dfrac{d\ln \left( {{x}^{2}}-4 \right)}{dt}\times \dfrac{dt}{dx}$
Taking derivative on both the sides of ${{x}^{2}}-4=t$ we get,
$2xdx=dt$
Dividing $dx$ on both the sides we get,
$2x=\dfrac{dt}{dx}$
Now, substituting ${{x}^{2}}-4=t$ and $2x=\dfrac{dt}{dx}$ in $\dfrac{d\ln \left( {{x}^{2}}-4 \right)}{dt}\times \dfrac{dt}{dx}$ we get,
$\begin{aligned} & \dfrac{d\ln \left( t \right)}{dt}\times 2x \\\ & =\dfrac{1}{t}\times 2x \\\ \end{aligned}$
Substituting $t={{x}^{2}}-4$ in the above we get,
$\dfrac{2x}{{{x}^{2}}-4}$