# How do you find (dy)/(dx) given sqrt(3x^7+y^2)=x?

Nov 16, 2016

${\sqrt{\left(3 {x}^{7} + {y}^{2}\right)}}^{2} = {x}^{2}$

$3 {x}^{7} + {y}^{2} = {x}^{2}$

$21 {x}^{6} + 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x$

$2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x - 21 {x}^{6}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x - 21 {x}^{6}}{2 y}$

However, since $x = 3 {x}^{7} + {y}^{2}$, we can substitute.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \sqrt{3 {x}^{7} + {y}^{2}} - 21 {x}^{6}}{2 y}$

Finally, we must note our restrictions on the variable. The √ can never be negative, so 3x^7 + y^2 ≥ 0. Also, $y \ne 0$.

Hopefully this helps!