Question

What is the derivative of ${{y}^{2}}=x$ when $y=\dfrac{1}{2}$?

Answer 1

Hint: The derivative of a function $y$ w.r.t $x$ is given as $\dfrac{dy}{dx}$.
In this case our function is ${{y}^{2}}$ therefore we have to apply chain rule to solve. The chain rule is applied when the function is composite i.e. of the form $f\left( g\left( x \right) \right)$ .
The derivative of ${{x}^{n}}$ is ${{x}^{n-1}}$.
We have to substitute the value of $y=\dfrac{1}{2}$.

Complete step by step answer:
Given,
${{y}^{2}}=x$
When $y=\dfrac{1}{2}$
To find the derivative of the function ${{y}^{2}}=x$ at $y=\dfrac{1}{2}$ .
${{y}^{2}}=x$ …(1)
Differentiate equation (1) w.r.t $x$ we get:
$\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( x \right)$ …(2)
We know that the differentiation of ${{x}^{n}}$ is ${{x}^{n-1}}$ and for the function we apply chain rule therefore applying these to equation (2) we get:
$2y\times \dfrac{dy}{dx}={{x}^{1-0}}$
$2y\dfrac{dy}{dx}=1$ …(3)
Now we can substitute the value of $y=\dfrac{1}{2}$ in equation (3) we get:
$2\times \dfrac{1}{2}\times \dfrac{dy}{dx}=1$ …(4)
On simplifying equation (4) we get:
$\dfrac{dy}{dx}=1$

Thus, the derivative of the function ${{y}^{2}}=x$ is $\dfrac{dy}{dx}=1$

Note: We can solve this question by differentiating w.r.t $y$ as it is not mentioned in the question we can use either of the methods.
When we differentiate the function ${{y}^{2}}=x$ w.r.t $y$ we get:
$\dfrac{d}{dy}\left( {{y}^{2}} \right)=\dfrac{d\left( x \right)}{dy}$
Now upon solving the above equation, we get:
$2y=\dfrac{dx}{dy}$
Now substituting the value of $y=\dfrac{1}{2}$ in the above equation we get:
$\begin{align}
  & 2\times \dfrac{1}{2}=\dfrac{dx}{dy} \\
 & \Rightarrow \dfrac{dx}{dy}=1 \\
\end{align}$
On reciprocating the above equation or cross multiplying the above equation we have:
$\dfrac{dy}{dx}=1$
Thus, we see that by using either of the methods the answer is the same therefore any method of your choice can be used.
Don’t get confused while applying the chain rule.
The chain rule for a composite function is given by:
$f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right).f\left( g'\left( x \right) \right)$
On correct application of chain rule the above answer could be obtained.