No, Not Possible

Calculus Level 4

$m$ and $n$ are constants. $y$ is defined in terms of $x$ such that

$\large (x+y)^{m+n}=x^my^n.$

Find $\dfrac{dy}{dx}$ .

\dfrac{my}{nx}

\dfrac{ny}{mx}

\dfrac yx

\dfrac mn

\dfrac xy

None of these choices

\dfrac nm

1 solution

Sravanth C.
May 28, 2016

Relevant wiki: Implicit Differentiation - Polynomials

Applying log on both sides of the equation and then finding the derivative is easier, so we get

$\begin{aligned} \dfrac{d}{dx}\log (x^my^n)&=\dfrac{d}{dx}\log (x+y)^{m+n}\\ \dfrac{d}{dx}m\log x + n\log y&=\dfrac{d}{dx}(m+n)\log (x+y)\\ \dfrac mx + \dfrac ny\dfrac{dy}{dx} &=\dfrac{(m+n)}{(x+y)}\left[1+\dfrac{dy}{dx}\right]\\ \end{aligned}$

Now bringing the dy/dx terms to one side and solving the expression we arrive at

$\begin{aligned} \dfrac{dy}{dx}\left[\dfrac ny-\dfrac{(m+n)}{(x+y)}\right]&=\dfrac{(m+n)}{(x+y)}-\dfrac mx\\ \dfrac{dy}{dx}\left[\dfrac{nx-my}{y(x+y)}\right]&=\dfrac{nx-my}{x(x+y)}\\ \dfrac{dy}{dx}&=\dfrac{nx-my}{x(x+y)}\times\dfrac{y(x+y)}{nx-my}\\ \dfrac{dy}{dx}&=\boxed{\dfrac yx} \end{aligned}$

Moderator note:

What happens when $nx - my = 0$ ?

Note the proper way of phrasing the question.

Calvin Lin Staff - 5 years ago

After taking log just apply df/dx=-((del f /del x)/(del f/del y)) where f is function of x, y and del stands for partial differentiation...... This makes the solution super easy

Samarth Agarwal - 5 years ago

No, Not Possible

1 solution

Moderator note:

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