# How do you find dy/dx by implicit differentiation of 1/2ycos(xy^2)?

##### 1 Answer
May 19, 2015

Finding $\frac{\mathrm{dy}}{\mathrm{dx}}$ by implicit differentiation involves solving an equation. There is no equation here, only an expression.

Here is how you differentiate with respect to $x$, assuming that $y$ is a function (or functions) of $x$:

To find $\frac{d}{\mathrm{dx}} \left(\frac{1}{2} y \cos \left(x {y}^{2}\right)\right)$, we'll need the product rule, the derivative of the cosine, the chain rule and the product rule some more.

I use the product rule in this order: $\left(f g\right) ' = f ' g + f g '$, so we write:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{2} y \cos \left(x {y}^{2}\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(\frac{1}{2} y\right)\right] \left[\cos \left(x {y}^{2}\right)\right] + \left[\frac{1}{2} y\right] \left[\frac{d}{\mathrm{dx}} \left(\cos \left(x {y}^{2}\right)\right)\right]$

Let's separate finding $\frac{d}{\mathrm{dx}} \left(\cos \left(x {y}^{2}\right)\right)$, because it may be challenging.

$\frac{d}{\mathrm{dx}} \left(\cos u\right) = - \sin u \frac{\mathrm{du}}{\mathrm{dx}}$ and in this case $u$ is a product of $x$ and the square of (the function of $x$ named $y$). Therefore, we get:

$\frac{d}{\mathrm{dx}} \left(\cos \left(x {y}^{2}\right)\right) = - \sin \left(x {y}^{2}\right) \left[\frac{d}{\mathrm{dx}} \left(x {y}^{2}\right)\right]$

$= - \sin \left(x {y}^{2}\right) \left[1 {y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

Returning to the main problem:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{2} y \cos \left(x {y}^{2}\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(\frac{1}{2} y\right)\right] \left[\cos \left(x {y}^{2}\right)\right] + \left[\frac{1}{2} y\right] \left[\frac{d}{\mathrm{dx}} \left(\cos \left(x {y}^{2}\right)\right)\right]$

$= \frac{1}{2} \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x {y}^{2}\right) + \frac{1}{2} y \left[- \sin \left(x {y}^{2}\right) \left(1 {y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]$

 = 1/2 cos(xy^2) dy/dx + 1/2y[-y^2sin(xy^2) - 2xysin(xy^2) dy/dx)]

 = 1/2 cos(xy^2) dy/dx - 1/2y^3sin(xy^2) - xy^2sin(xy^2) dy/dx)

More algebraic rewriting is possible, but we cannot solve for anything because there is no equation.