Question

Differentiate the equation \[x\sqrt x + y\sqrt y = a\sqrt a \] with respect to x.

Answer 1

Hint: Here we write the value of square root as power \[\dfrac{1}{2}\] for each term in the square root and using property of exponents that \[{p^m} \times {p^q} = {p^{m + q}}\] we combine the powers of each term. We differentiate each term separately where y can be a function of x so we use chain rule in differentiation of y and term on RHS is constant term.
* \[\sqrt[n]{p} = {(p)^{\dfrac{1}{n}}}\]
* We have the method of differentiation \[\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}\], where differentiation of constants is zero.
* Chain rule of differentiation: \[\dfrac{d}{{dx}}[f(g(x)] = f'(g(x)) \times g'(x)\], where \[f'(g(x))\] is differentiation of f with respect to x and \[g'(x)\] is differentiation of g with respect to x.

Complete step-by-step answer:
We have an equation \[x\sqrt x + y\sqrt y = a\sqrt a \]
Since we know that \[\sqrt[n]{p} = {(p)^{\dfrac{1}{n}}}\]. Here the value of n is 2 for all the terms.
\[ \Rightarrow x \times {(x)^{\dfrac{1}{2}}} + y \times {(y)^{\dfrac{1}{2}}} = a \times {(a)^{\dfrac{1}{2}}}\]
Use the property that when the base is the same, powers can be added.
\[ \Rightarrow {x^{1 + \dfrac{1}{2}}} + {y^{1 + \dfrac{1}{2}}} = {a^{1 + \dfrac{1}{2}}}\]
Take LCM in the power of each term
\[ \Rightarrow {x^{\dfrac{{2 + 1}}{2}}} + {y^{\dfrac{{2 + 1}}{2}}} = {a^{\dfrac{{2 + 1}}{2}}}\]
\[ \Rightarrow {x^{\dfrac{3}{2}}} + {y^{\dfrac{3}{2}}} = {a^{\dfrac{3}{2}}}\]
Now we differentiate both sides of the equation.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{x^{\dfrac{3}{2}}} + {y^{\dfrac{3}{2}}}} \right) = \dfrac{d}{{dx}}\left( {{a^{\dfrac{3}{2}}}} \right)\]
Break the differentiation on LHS of the equation for separate terms.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{x^{\dfrac{3}{2}}}} \right) + \dfrac{d}{{dx}}\left( {{y^{\dfrac{3}{2}}}} \right) = \dfrac{d}{{dx}}\left( {{a^{\dfrac{3}{2}}}} \right)\]
We know that \[\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}\], y is a function of x so we apply chain rule to differentiation of y and differentiation of constant term is 0.
\[ \Rightarrow \dfrac{3}{2}{(x)^{\dfrac{3}{2} - 1}} + \dfrac{3}{2}{(y)^{\dfrac{3}{2} - 1}} \times \dfrac{{dy}}{{dx}} = 0\]
Take LCM in power of each term and take the constant term common.
\[ \Rightarrow \dfrac{3}{2}[{(x)^{\dfrac{{3 - 2}}{2}}} + \dfrac{3}{2}{(y)^{\dfrac{{3 - 2}}{2}}} \times \dfrac{{dy}}{{dx}}] = 0\]
Cross multiply the constant term to RHS of the equation
\[ \Rightarrow {(x)^{\dfrac{1}{2}}} + {(y)^{\dfrac{1}{2}}} \times \dfrac{{dy}}{{dx}} = 0\]
Since, \[\sqrt p = {(p)^{\dfrac{1}{2}}}\]
\[ \Rightarrow \sqrt x + \sqrt y \dfrac{{dy}}{{dx}} = 0\]
Thus differentiation of \[x\sqrt x + y\sqrt y = a\sqrt a \]with respect to x is \[\sqrt x + \sqrt y \dfrac{{dy}}{{dx}} = 0\]
We can further solve the equation by shifting the term with x on one side of the equation.
\[ \Rightarrow \sqrt y \dfrac{{dy}}{{dx}} = - \sqrt x \]
Dividing both sides by \[\sqrt y \]
\[ \Rightarrow \dfrac{{\sqrt y }}{{\sqrt y }}\dfrac{{dy}}{{dx}} = - \dfrac{{\sqrt x }}{{\sqrt y }}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \sqrt {\dfrac{x}{y}} \]

Note: Students are likely to make the mistake of not differentiating y with respect to x because they think it is not mentioned anywhere in the question so y will not be a function of x. Keep in mind the sign of value changes from positive to negative and vice versa when shifted from one side of the equation to the other side of the equation.