Question

If ${y^2} = 4x$, find the rate at which y is changing with respect to x, when $x = 4$

Answer 1

Hint : Here the question is related to the differentiation. The sentence rate of change means we have to find the derivative. The function is a power function so first we apply square root and then we differentiate the function and finally we substitute the value of x as 4 hence we obtain the result

Complete step by step solution:
A function is a process or a relation that associates each element x of a set X, the domain of the function, to a single element y of another set Y (possibly the same set), the codomain of the function.
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument.
Now consider the function which is given in the question
$\Rightarrow {y^2} = 4x$
The function y is a power function. Taking square root on both sides we write it as
$\Rightarrow y = \sqrt {4x}$
On simplifying we get
$\Rightarrow y = \pm 2\sqrt x$
So we get two functions
1. $y = 2\sqrt x$
On differentiating the above function with respect to x we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 2\dfrac{1}{{2\sqrt x }}$
On simplifying we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt x }}$
On substituting the value of x we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt 4 }}$
On taking square root we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \pm \dfrac{1}{2}$
Now we consider another function
2. $y = - 2\sqrt x$
On differentiating the above function with respect to x we get
$\Rightarrow \dfrac{{dy}}{{dx}} = - 2\dfrac{1}{{2\sqrt x }}$
On simplifying we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{\sqrt x }}$
On substituting the value of x we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{\sqrt 4 }}$
On taking square root we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \mp \dfrac{1}{2}$
Hence we have determined the rate of change of y with respect to x.
So, the correct answer is “$\mp \dfrac{1}{2}$”.

Note : Here in this function without taking square root we can directly differentiate the function, but it is not the appropriate way. When we differentiate the function directly without applying the square root of the function, we have a dependent variable y. So it’s better to apply the square root initially.