Question

Find the value of $\dfrac{dx}{dy}$ at y=1, if $\sqrt{x}+\sqrt{y}=4$.

Answer 1

Hint: The derivative of $\sqrt{x}$ is given as $\dfrac{d(\sqrt{x})}{dx}=\dfrac{1}{2\sqrt{x}}$. Using this we can find out the solution of this question.
Complete step-by-step answer:
Here the given equation $\sqrt{x}+\sqrt{y}=4$
Taking $\sqrt{y}$ to the RHS of the equation, we get
$\sqrt{x}=4-\sqrt{y}$.
Hence, we get a function $x$ in terms of variable $y$, or we can say $\sqrt{x}=f(y)$.
Now, we will differentiate both sides of the equation with respect to $y$.
On differentiating both sides of the equation with respect to $y$, we get,
$\dfrac{1}{2\sqrt{x}}.\dfrac{dx}{dy}=\dfrac{-1}{2\sqrt{y}}$
So, $\dfrac{dx}{dy}=\dfrac{-1}{2\sqrt{y}}\div \dfrac{1}{2\sqrt{x}}$
Now, in the question it is given that $\sqrt{x}+\sqrt{y}=4$.
So, at $y=1$, we get $\sqrt{x}+\sqrt{1}=4$
$\Rightarrow$the value of $\sqrt{x}=4\pm 1$ and the value of$\sqrt{y}=\pm 1$
Hence, we have four cases:
Case 1: $\sqrt{x}=4-1=3$ and $\sqrt{y}=1$
Now, we will substitute $\sqrt{x}=3$ and $\sqrt{y}=1$ in the expression of $\dfrac{dx}{dy}$.
On substituting $\sqrt{x}=3$and$\sqrt{y}=1$ in the expression of $\dfrac{dx}{dy}$, we get,
${{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(3)}$
$=\dfrac{-1}{2}\div \dfrac{1}{6}$

& =\dfrac{-1}{2}\times 6 \\\ & =-3 \\\ \end{aligned}$$ Case 2: $$\sqrt{x}=4-1=3$$ and $$\sqrt{y}=-1$$ Now, we will substitute $$\sqrt{x}=3$$ and $$\sqrt{y}=-1$$ in the expression of $$\dfrac{dx}{dy}$$. On substituting $$\sqrt{x}=3$$ and $$\sqrt{y}=-1$$ in the expression of $$\dfrac{dx}{dy}$$, we get, $${{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(3)}$$ $$=\dfrac{-1}{-2}\div \dfrac{1}{6}$$ $$\begin{aligned} & =\dfrac{1}{2}\times 6 \\\ & =3 \\\ \end{aligned}$$ Case 3:$$\sqrt{x}=4+1=5$$ and $$\sqrt{y}=1$$ Now, we will substitute $$\sqrt{x}=5$$ and $$\sqrt{y}=1$$ in the expression of $$\dfrac{dx}{dy}$$. On substituting $$\sqrt{x}=5$$ and $$\sqrt{y}=1$$ in the expression of $$\dfrac{dx}{dy}$$, we get, $${{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(5)}$$ $$=\dfrac{-1}{2}\div \dfrac{1}{10}$$ $$\begin{aligned} & =\dfrac{-1}{2}\times 10 \\\ & =-5 \\\ \end{aligned}$$ Case 4:$$\sqrt{x}=4+1=5$$ and $$\sqrt{y}=-1$$ Now, we will substitute $$\sqrt{x}=5$$ and $$\sqrt{y}=-1$$ in the expression of $$\dfrac{dx}{dy}$$. On substituting $$\sqrt{x}=5$$ and $$\sqrt{y}=-1$$ in the expression of $$\dfrac{dx}{dy}$$, we get, $${{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(5)}$$ $$=\dfrac{-1}{-2}\div \dfrac{1}{10}$$ $$\begin{aligned} & =\dfrac{1}{2}\times 10 \\\ & =5 \\\ \end{aligned}$$ Hence, the value of the derivative $$\dfrac{dx}{dy}$$ is equal to $$(-3)$$, $$3,(-5)$$ and $$5$$. Note: The question has asked to find the value of $$\dfrac{dx}{dy}$$ and not $$\dfrac{dy}{dx}$$. The two can often be confused. $$\dfrac{dy}{dx}$$ is the derivative of function $$y$$ with respect to variable $$x$$ whereas $$\dfrac{dx}{dy}$$ is the derivative of function $$x$$ with respect to variable $$y$$. The value of both is not the same. So, students should be careful in such cases.