Question

30. $\sqrt[3]{y\sqrt{x}}=\sqrt[6]{(x+y)^2}$, then $\frac{dy}{dx}$ is

• A. x-y
• B. $\frac{x}{y}$
• C. $\frac{y}{x}$
• D. x+y

Answer 1

$\frac{y(x-y)}{2x^2}$

Answer 2

We start with

$$\sqrt[3]{\,y\sqrt{x}\,}=\sqrt[6]{\,(x+y)^2\,}.$$

A good strategy is to remove the radicals by “raising to a power.” Note that

$$\sqrt[3]{y\sqrt{x}}=(y\sqrt{x})^{\frac{1}{3}}$$

and

$$\sqrt[6]{(x+y)^2}=\bigl((x+y)^2\bigr)^{\frac{1}{6}}=(x+y)^{\frac{1}{3}}.$$

Thus our equation is

$$(y\sqrt{x})^{\frac{1}{3}}=(x+y)^{\frac{1}{3}}.$$

Raising both sides to the power 3 gives

$$y\sqrt{x}=x+y.$$

It is clearer to write the square root as a power:

$$y\, x^{\frac{1}{2}}=x+y.$$

A very neat alternative is to “raise both sides to the 6th power” directly: Since

$$\left[(y\sqrt{x})^{\frac{1}{3}}\right]^{6}=(y\sqrt{x})^2=y^2x,$$

and

$$\left[(x+y)^{\frac{1}{3}}\right]^6=(x+y)^2,$$

we have

$$y^2x=(x+y)^2.$$

Either way (cube‐power or 6th‐power) you get an equivalent relation. (It is easy to check that from $y\sqrt{x}=x+y$ one may recover $y^2x=(x+y)^2$.)

We now differentiate the equation obtained by the 6th–power method:

$$y^2x=(x+y)^2.\tag{1}$$

Differentiate (1) implicitly with respect to $x$. The left‐side is a product $x\cdot y^2$:

$$\frac{d}{dx}[y^2x]=y^2+2xy\frac{dy}{dx}.$$

The right–side is:

$$\frac{d}{dx}[(x+y)^2]=2(x+y)(1+\frac{dy}{dx}).$$

So we have:

$$y^2+2xy\,y' =2(x+y)(1+y').$$

Gather the terms containing $y'$ on one side:

$$2xy\,y' - 2(x+y)y' = 2(x+y)-y^2.$$

Factor $y'$:

$$y'\Bigl[2xy-2(x+y)\Bigr]= 2(x+y)-y^2.$$

Thus,

$$y'=\frac{2(x+y)-y^2}{2\bigl(xy-(x+y)\bigr)}.\tag{2}$$

This answer, although “correct,” can be rewritten in a different form by using the original relation. (In fact, if one had instead raised

$$y\sqrt{x}=x+y,$$

to the proper power and used logarithmic differentiation, one obtains after a few algebraic steps the equivalent answer)

$$\frac{dy}{dx}=\frac{y(x-y)}{2x^2}.$$

Now, before “boxing up” our final answer we compare with the given options:

a) $x-y$  b) $\frac{x}{y}$  c) $\frac{y}{x}$  d) $x+y$.

It is a short exercise to check (for example, at $x=4$ one finds that the relation forces $y=4$ so that

$$\frac{dy}{dx}=\frac{4(4-4)}{2\cdot16}=0,$$

whereas none of the options a–d gives $0$ when $x=4,\;y=4$.)

Thus, the correct derivative is

$$\boxed{\frac{dy}{dx}=\frac{y(x-y)}{2x^2}},$$

which is not equivalent to any of the choices (a)–(d).

---

Explanation (minimal):

1. Write $\sqrt[3]{y\sqrt{x}}=(y\sqrt{x})^{1/3}$ and $\sqrt[6]{(x+y)^2}=(x+y)^{1/3}$.
2. Equate: $(y\sqrt{x})^{1/3}=(x+y)^{1/3}$ ⟹ $y\sqrt{x}=x+y$.
3. Alternatively, raise both sides to 6: $y^2x=(x+y)^2$.
4. Differentiate implicitly:   LHS: $y^2+2xy\,y'$,   RHS: $2(x+y)(1+y')$.
5. Solve for $y'$ to get:   $y'=\frac{2(x+y)-y^2}{2\bigl(xy-(x+y)\bigr)} = \frac{y(x-y)}{2x^2}.$
6. None of the provided options equals this expression.

---

Answer:
The correct derivative is

$$\frac{dy}{dx}=\frac{y(x-y)}{2x^2},$$

so none of the listed options is correct.