y = cos³x tan x

The right hand side is a product of (cos x)³ and (tan x).

Now (cos x)³ is a power of a function and so we use Differentiating Powers of a Function:

`d/(dx)u^3=3u^2(du)/(dx)`

With u = cos x, we have:

`d/(dx)(cos\ x)^3=3(cos\ x)^2(-sin\ x)`

Now, from our rules above, we have:

`d/(dx)tan\ x=sec^2x`

Using the Product Rule and Properties of tan x, we have:

` {: ((dy)/(dx),=[cos^3x\ sec^2x]+tan\ x[3(cos\ x)^2(-sin\ x)]),(,=(cos^3x)/(cos^2x)+(sin\ x)/(cos\ x)[3(cos\ x)^2(-sin\ x)]),(,=cos\ x-3\ sin^2x\ cos\ x) :}`

We need to determine if this expression creates a true statement when we substitute it into the LHS of the equation given in the question.

` {: (LHS=cos\ x(dy)/(dx)+3y\ sin\ x-cos^2x),(=cos\ x(cos\ x-3\ sin^2x\ cos\ x)+3(cos^3x\ tan\ x)sin\ x-cos^2x),(=cos^2x-3\ sin^2x\ cos^2x+3\ sin^2x\ cos^2x-cos^2x),(=0),(=RHS) :}`

We have shown that it is true.