Question

If $x^3+8xy+y^3=64$, then $\frac{dy}{dx}=$

• A. $\frac{-(3x^2+8y)}{(8x+3y^2)}$
• B. $\frac{(3x^2+8y)}{(8x+3y^2)}$
• C. $\frac{(3x+8y)}{(8x^2+3y)}$
• D. None of these

Answer 1

The correct option is A

$\frac{-(3x^2+8y)}{(8x+3y^2)}$

Explanation for the correct option.

$x^3+8xy+y^3=64$

Differentiating both sides w.r.t $x$, we get

$\begin{array}{rcl}3x^2+8x\left(\frac{dy}{dx}\right)+8y+3y^2\left(\frac{dy}{dx}\right)& =& 0\\ \Rightarrow \left(8x+3y^2\right)\frac{dy}{dx}& =& -\left(3x^2+8y\right)\\ \Rightarrow \frac{dy}{dx}& =& \frac{-\left(3x^2+8y\right)}{\left(8x+3y^2\right)}\end{array}$

Hence, option A is correct.