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,
\[{{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(3)}\]
\[=\dfrac{-1}{2}\div \dfrac{1}{6}\]
\[\begin{align}
  & =\dfrac{-1}{2}\times 6 \\
 & =-3 \\
\end{align}\]
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,
\[{{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(3)}\]
\[=\dfrac{-1}{-2}\div \dfrac{1}{6}\]
\[\begin{align}
  & =\dfrac{1}{2}\times 6 \\
 & =3 \\
\end{align}\]
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,
\[{{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(5)}\]
\[=\dfrac{-1}{2}\div \dfrac{1}{10}\]
\[\begin{align}
  & =\dfrac{-1}{2}\times 10 \\
 & =-5 \\
\end{align}\]
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,
\[{{(\dfrac{dx}{dy})}_{y=1}}=\dfrac{-1}{2\sqrt{1}}\div \dfrac{1}{2(5)}\]
\[=\dfrac{-1}{-2}\div \dfrac{1}{10}\]
\[\begin{align}
  & =\dfrac{1}{2}\times 10 \\
 & =5 \\
\end{align}\]
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.