3
$\begingroup$

If $x^{13}y^{7}=(x+y)^{20}$ , then $\frac{dy}{dx}$

directly doing it makes it very complicated so, I did this $\left(\frac{x}{y}\right)^{13}=\left(1+\frac{x}{y} \right)^{20}$. following are the options for solution (a) $\frac{y^2}{x^2}$ (b)$\frac{x^2}{y^2}$ (c)$\frac{x}{y}$ (d)$\frac{y}{x}$ thanks for any hints.

$\endgroup$
3
  • $\begingroup$ Is the answer D? $\endgroup$ Mar 24, 2016 at 16:52
  • $\begingroup$ yez, how did you conclude that? $\endgroup$
    – Onix
    Mar 24, 2016 at 16:57
  • $\begingroup$ I posted a solution. It is pretty long , so I am thinking of a shorter solution. $\endgroup$ Mar 24, 2016 at 17:04

1 Answer 1

1
$\begingroup$

I differentiated directly,

$$13x^{12}y^7+7x^{13}y^6y'=20(x+y)^{19}(1+y')$$

$${13x^{13}y^7\over {x}}+{7x^{13}y^7y'\over y}={20(x+y)^{20}(1+y')\over {(x+y)}}$$

$$(x+y)(13y+7xy')=20xy(1+y')$$

Cancelling $x^{13}y^7$ and rearranging we get,

$$y'={13y^2-7xy\over {13xy-7x^2}}$$

Divide the numerator and the dinominator by$x^2$

$$y'={13({y\over x})^2-7({y\over x})\over {13({y\over x})-7}}$$

$$y'={y\over x}$$

$\endgroup$
2
  • $\begingroup$ thanks @Dhanush Krishna I missed using the equation in the differentiation really intuitive thanks. $\endgroup$
    – Onix
    Mar 24, 2016 at 17:09
  • $\begingroup$ @Abomm: welcome $\endgroup$ Mar 25, 2016 at 2:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .