# How do you differentiate implicitly to find the slope of the curve y^4 + x^3 = y^2 + 10x at the given point (0,1)?

May 18, 2015

${y}^{4} + {x}^{3} = {y}^{2} + 10 x$

Differentiate both sides of the equation with respect to $x$.

Use whichever notation you prefer:

${D}_{x} \left({y}^{4} + {x}^{3}\right) = {D}_{x} \left({y}^{2} + 10 x\right)$

$\frac{d}{\mathrm{dx}} \left({y}^{4} + {x}^{3}\right) = \frac{d}{\mathrm{dx}} \left({y}^{2} + 10 x\right)$

$4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {x}^{2} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 10$

Now, if we want the general formula for $\frac{\mathrm{dy}}{\mathrm{dx}}$, solve algebraically for $\frac{\mathrm{dy}}{\mathrm{dx}}$, but all we have been aksed for is $\frac{\mathrm{dy}}{\mathrm{dx}}$ when $x = 0$ and $y = 1$, so let's just do that:

$4 {\left(1\right)}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {\left(0\right)}^{2} = 2 \left(1\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 10$

$4 \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \frac{\mathrm{dy}}{\mathrm{dx}} + 10$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{10}{2} = 5$