Question

How do you differentiate $y=\dfrac{3x}{{{x}^{2}}-3}$?

Answer 1

Hint: Now to differentiate the given function we will use the property of differentiation ${{\left( \dfrac{f}{g} \right)}^{'}}=\dfrac{f'g-g'f}{{{g}^{2}}}$ . Now we know that $\dfrac{d\left( f\pm g \right)}{dx}=\dfrac{df}{dx}\pm \dfrac{dg}{dx}$, $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d\left( c \right)}{dx}=0$ . Hence using this we will solve differentiation of the function. Now we will further simplify the expression by using formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$

Complete step-by-step answer:
Now we are given with a function which is in fraction. To differentiate the given function we will have to use the division rule for fraction.
Now let us first understand the properties of differentiation.
Now differentiation of function in addition or subtraction is given by addition or subtraction of differentiation of the function.
Hence we have $\dfrac{d\left( f\pm g \right)}{dx}=\dfrac{df}{dx}\pm \dfrac{dg}{dx}$ .
Now for functions in multiplication we have $\left( f.g \right)'=f'g+g'f$ . Where $\left( f \right)'$ represents differentiation of the function f.
Similarly for function in division form we have, ${{\left( \dfrac{f}{g} \right)}^{'}}=\dfrac{f'g-g'f}{{{g}^{2}}}$
Now the given function is in the form $\dfrac{f}{g}$ where $f=3x$ and $g={{x}^{2}}-3$ .
Now we know that the differentiation of such function is given by,
$\Rightarrow \dfrac{\dfrac{d\left( 3x \right)}{dx}\left( {{x}^{2}}-3 \right)-\dfrac{d\left( {{x}^{2}}-3 \right)}{dx}\left( 3x \right)}{{{\left( {{x}^{2}}-3 \right)}^{2}}}$
Now we know that the differentiation of the function of the form ${{x}^{n}}$ is given by $n{{x}^{n-1}}$ and $\dfrac{d\left( cf \right)}{df}=c\dfrac{df}{df}$ Hence we have $\dfrac{d\left( 3x \right)}{dx}=3$
Now similarly we have $\dfrac{d\left( f+g \right)}{dx}=\dfrac{d\left( f \right)}{dx}-\dfrac{d\left( g \right)}{dx}$ and we know that $\dfrac{d\left( c \right)}{dx}=0$
Hence we get, $\dfrac{d\left( {{x}^{2}}-3 \right)}{dx}=2x$.
Hence substituting these values we get,
$\Rightarrow \dfrac{3\left( {{x}^{2}}-3 \right)-2x\left( 3x \right)}{{{\left( {{x}^{2}}-3 \right)}^{2}}}$
Now we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. Hence we get,
$\Rightarrow \dfrac{3{{x}^{2}}-9-6{{x}^{2}}}{{{x}^{4}}-6{{x}^{2}}+9}$
Now on simplifying the expression we get,
$\Rightarrow \dfrac{-3{{x}^{2}}-9}{{{x}^{4}}-6{{x}^{2}}+9}$
Hence the differentiation of the given function is given by $\dfrac{-3{{x}^{2}}-9}{{{x}^{4}}-6{{x}^{2}}+9}$ .

Note: Now note that for differentiation we have ${{\left( \dfrac{f}{g} \right)}^{'}}=\dfrac{f'g-g'f}{{{g}^{2}}}$ . Note that there is no differentiation term in denominator and since there is subtraction in numerator keep a note of order of the terms. Also note that differentiation of constant is 0.