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# How to find the derivative of $\sqrt x + \sqrt y = 1$ by using the implicit differentiation?

Last updated date: 02nd Aug 2024
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Hint:
Here we will differentiate the given equation to solve the question. First, we will write the equation in the exponent form of the variables. Then we will differentiate the equation and after differentiating we will modify the equation to get the value of $\dfrac{{dy}}{{dx}}$.

Complete step by step solution:
The given equation is $\sqrt x + \sqrt y = 1$.
First, we will write the given equation in the exponent form of the variables. Therefore we can write the given equation as
$\Rightarrow {x^{\dfrac{1}{2}}} + {y^{\dfrac{1}{2}}} = 1$
Now we will simply differentiate the above equation to get the value of $\dfrac{{dy}}{{dx}}$. Therefore, we get
$\Rightarrow \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}} + {y^{\dfrac{1}{2}}}} \right) = \dfrac{d}{{dx}}\left( 1 \right)$
We know that the differentiation of any constant value is equal to zero. Therefore, we get
$\Rightarrow \dfrac{1}{2}{x^{\dfrac{1}{2} - 1}} + \dfrac{1}{2}{y^{\dfrac{1}{2} - 1}} \cdot \dfrac{{dy}}{{dx}} = 0$
Now we will simplify and solve the above equation, we get
$\Rightarrow \dfrac{1}{2}{x^{ - \dfrac{1}{2}}} + \dfrac{1}{2}{y^{ - \dfrac{1}{2}}} \cdot \dfrac{{dy}}{{dx}} = 0$
We know that we can write ${x^{ - \dfrac{1}{2}}}$ as $\dfrac{1}{{\sqrt x }}$ and ${y^{ - \dfrac{1}{2}}}$ as $\dfrac{1}{{\sqrt y }}$. Therefore putting these values in the equation we get
$\Rightarrow \dfrac{1}{{2\sqrt x }} + \dfrac{1}{{2\sqrt y }} \cdot \dfrac{{dy}}{{dx}} = 0$
Now we will modify the above equation to get the value of $\dfrac{{dy}}{{dx}}$. Therefore, we get
$\Rightarrow \dfrac{1}{{2\sqrt y }} \cdot \dfrac{{dy}}{{dx}} = - \dfrac{1}{{2\sqrt x }}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{2\sqrt y }}{{2\sqrt x }}$
Now we will cancel out the common terms in the numerator and the denominator. Therefore, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sqrt y }}{{\sqrt x }}$

Hence the derivative of $\sqrt x + \sqrt y = 1$ by using the implicit differentiation is equal to $\dfrac{{dy}}{{dx}} = - \dfrac{{\sqrt y }}{{\sqrt x }}$.

Note:
A differentiable function may be defined as is a function whose derivative exists at every point in its range of domain. We should remember that a differentiable function is always continuous but the converse is not true which means a function may be continuous but not always differentiable. In this type of question we should simplify the equation in terms of the values given in the question.
We should know the basic formula of the differentiation of the $uv$ and the formula of the differentiation of the $\dfrac{u}{v}$,
$\begin{array}{l} \dfrac{d}{{dx}}\left( {uv} \right) = uv' + u'v\\ \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}} \end{array}$