Question

How do you find the derivative of $\dfrac{x}{{{x^2} - 4}}$ ?

Answer 1

Hint: We can find the derivative of the given expression simply by applying the Quotient Rule of Derivative and simplifying it further.

Formula used:
Quotient Rule: $y' = \dfrac{{g(x) \times f'(x) - f(x) \times g'(x)}}{{{{(g(x))}^2}}}$

Complete step by step answer:
We will find the derivative of the given expression using Quotient Rule:
Explaining Quotient Rule:
Suppose we have:
$ \Rightarrow y = \dfrac{{f(x)}}{{g(x)}}$
Then, using the Quotient Rule:
$ \Rightarrow y' = \dfrac{{g(x)f'(x) - f(x) \times g'(x)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$
In simple words, you take the derivative of f(x) multiplied by g(x), subtract f(x) multiplied by the derivative of g(x) and divide all that by ${\left( {g\left( x \right)} \right)^2}$.
Start by rewriting the expression:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right)$
Simplify by using Quotient Rule explained above:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = \dfrac{{\left( {\left( {{x^2} - 4} \right)\left( 1 \right)} \right) - \left( {\left( x \right)\left( {2x} \right)} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}}$
Simplify:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = \dfrac{{\left( {\left( {{x^2} - 4} \right)\left( 1 \right)} \right) - \left( {\left( x \right)\left( {2x} \right)} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = \dfrac{{\left( {{x^2} - 4} \right) - 2{x^2}}}{{{{\left( {{x^2} - 4} \right)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = - \dfrac{{{x^2} + 4}}{{{{\left( {{x^2} - 4} \right)}^2}}}$

So the derivative of $\dfrac{x}{{{x^2} - 4}}$ is $ - \dfrac{{{x^2} + 4}}{{{{\left( {{x^2} - 4} \right)}^2}}}$.

Note: The alternative method to find the derivative is by using product, power and chain rule.
Explanation:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right)$
Solve by taking the inverse of the denominator:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = \dfrac{d}{{dx}}x{\left( {{x^2} - 4} \right)^{ - 1}}$
Simplify using Product Rule:
Explaining Product rule:
Suppose we have:
$ \Rightarrow y = f(x) \times g(x)$
Then, using the Product rule:
$ \Rightarrow y' = f(x) \times g'(x) + f'(x) \times g(x)$
In simple words, keep the first term as it is and differentiate the second term, then differentiate the first term and keep the second term as it is or vice-versa.
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = \left( {\dfrac{d}{{dx}}x} \right){\left( {{x^2} - 4} \right)^{ - 1}} + x\left( {\dfrac{d}{{dx}}{{\left( {{x^2} - 4} \right)}^{ - 1}}} \right)$
After derivation, we get
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = {\left( {{x^2} - 4} \right)^{ - 1}} - 2{x^2}{\left( {{x^2} - 4} \right)^{ - 2}}$
Now, using Power Rule and Chain Rule to simplify the above expression:
Explaining Power rule:
Suppose we have:
$ \Rightarrow y = a{x^n}$
Then using Power rule:
$ \Rightarrow y' = (a \times n){x^{n - 1}}$
In simple words, multiply the variable’s exponent n, by its coefficient a, then subtract 1 from the exponent. If there’s no coefficient (the coefficient is 1), then the exponent will become the new coefficient.
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = \dfrac{{\left( {{x^2} - 4} \right) - 2{x^2}}}{{{{\left( {{x^2} - 4} \right)}^2}}}$
Simplify:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{{x^2} - 4}}} \right) = - \dfrac{{{x^2} + 4}}{{{{\left( {{x^2} - 4} \right)}^2}}}$
So the derivative of $\dfrac{x}{{{x^2} - 4}}$ is $ - \dfrac{{{x^2} + 4}}{{{{\left( {{x^2} - 4} \right)}^2}}}$.
We will get the same result with both the methods but this method is quite long.