Question

How do you find the derivative of $y = \dfrac{2}{{3{x^2}}}$?

Answer 1

Derivative of a function measures the rate of change of the function value with respect to change in the argument value. To find the derivative of the given function we will first bring the ${x^2}$ term to the numerator and then use the formula for differentiation of ${x^2}$ with respect to $x$.
Formula used:
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$

Complete step-by-step solution:
Given function:
$y = \dfrac{2}{{3{x^2}}}$
Bringing ${x^2}$ to the numerator we get;
$\Rightarrow y = \dfrac{{2{x^{ - 2}}}}{3}$
Now we will differentiate both sides with respect to $x$. So, we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{3}\dfrac{{d{x^{ - 2}}}}{{dx}}$
Now we will use the formula that: $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{3} \times \left( { - 2} \right){x^{ - 2 - 1}}$
On further simplification we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 4}}{3}{x^{ - 3}}$
On further simplification by shifting the ${x^{ - 3}}$ to the denominator we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 4}}{{3{x^3}}}$
Additional Information:
A function $y = f\left( x \right)$ is differentiable at a point $a$ of its domain, if its domain contains an open interval $I$ and the limit,
$L = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {a + h} \right) - f\left( a \right)}}{h}$ , exists.
If the function $f$ is differentiable at $a$, that is if the limit $L$ exists, then this limit is called the derivative of $f$ at $a$. The process of finding a derivative is called differentiation. The inverse process of differentiation is called integration. Derivative of a function at a point gives the slope of the tangent to the function at that function. For a linear function this slope is constant and it does not change with the point because the derivative of a linear function is constant.

Note: We can also solve this question by the product rule of differentiation. For that we will first shift the ${x^2}$ to the LHS then differentiate both sides and apply the product rule of differentiation.
Given;
$y = \dfrac{2}{{3{x^2}}}$
On shifting we get,
$\Rightarrow {x^2}y = \dfrac{2}{3}$
Now we will differentiate both sides with respect to $x$.
$\Rightarrow \dfrac{{d\left( {{x^2}y} \right)}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{2}{3}} \right)$
RHS will become zero because differentiation of constant is zero
$\Rightarrow \dfrac{{d\left( {{x^2}y} \right)}}{{dx}} = 0$
Differentiating using the product rule we get;
$\Rightarrow {x^2}\dfrac{{dy}}{{dx}} + y\dfrac{{d{x^2}}}{{dx}} = 0$
On solving we get;
$\Rightarrow {x^2}\dfrac{{dy}}{{dx}} + 2xy = 0$
On shifting the terms, we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2xy}}{{{x^2}}}$
Now we will put the value of $y$. So, we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2x}}{{{x^2}}} \times \dfrac{2}{{3{x^2}}}$
On solving by cancelling the terms we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 4}}{{3{x^3}}}$