Question

Find dy/dx by implicit differentiation. 5y sin(x2) = 3x sin(y2)

Answer 1

Assuming the equation reads

$5y \sin(x^2) = 3x \sin(y^2)$

Differentiating both sides with respect to x gives

$\dfrac{\mathrm d}{\mathrm dx}\left[5y \sin(x^2)\right] = \dfrac{\mathrm d}{\mathrm dx}\left[3x \sin(y^2)\right] \\\\ 5 \dfrac{\mathrm d}{\mathrm dx}\left[y \sin(x^2)\right] = 3 \dfrac{\mathrm d}{\mathrm dx}\left[x \sin(y^2)\right]$

By the product rule,

$5 \left(\dfrac{\mathrm d}{\mathrm dx}[y] \sin(x^2) + y \dfrac{\mathrm d}{\mathrm dx}\left[\sin(x^2)\right]\right) = 3 \left(\dfrac{\mathrm d}{\mathrm dx}[x] \sin(y^2) + x \dfrac{\mathrm d}{\mathrm dx}\left[\sin(y^2)\right]\right)$

By the chain rule,

$5 \left(\sin(x^2) \dfrac{\mathrm dy}{\mathrm dx} + y \cos(x^2) \dfrac{\mathrm d}{\mathrm dx}\left[x^2\right]\right) = 3 \left(\sin(y^2) + x \cos(y^2) \dfrac{\mathrm d}{\mathrm dx}\left[y^2\right]\right) \\\\ 5 \left(\sin(x^2) \dfrac{\mathrm dy}{\mathrm dx} + 2xy \cos(x^2)\right) = 3 \left(\sin(y^2) + 2xy \cos(y^2) \dfrac{\mathrm dy}{\mathrm dx}\right)$

Solve for dy/dx :

$5\sin(x^2) \dfrac{\mathrm dy}{\mathrm dx} + 10 xy \cos(x^2)\right) = 3 \sin(y^2) + 6xy \cos(y^2) \dfrac{\mathrm dy}{\mathrm dx} \\\\ \left(5\sin(x^2)-6xy\cos(y^2)\right) \dfrac{\mathrm dy}{\mathrm dx} = 3 \sin(y^2) - 10 xy \cos(x^2) \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \boxed{\dfrac{3 \sin(y^2) - 10 xy \cos(x^2)}{5\sin(x^2)-6xy\cos(y^2)}}$