First, let: `u = x^2+ 3` and so `y = sin u`.

We have:

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=cos u(du)/(dx)`

`=cos(x^2+3)(d(x^2+3))/(dx)`

`=2x\ cos(x^2+3)`

IMPORTANT:

cos x² + 3

does not equal

cos(x² + 3).

The brackets make a big difference. Many students have trouble with this.

Here are the graphs of y = cos x² + 3 (in green) and y = cos(x² + 3) (shown in blue).

The first one, y = cos x² + 3, or y = (cos x²) + 3, means take the curve y = cos x² and move it up by `3` units.

The second one, y = cos(x² + 3), means find the value (x² + 3) first, then find the cosine of the result.

They are quite different!

Let u = 3x⁴ and so `y = cos u`.

Then

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=-sin u(du)/(dx)`

`=-sin(3x^4)(d(3x^4))/(dx)`

`=-12x^3sin 3x^4`

This example has a function of a function of a function.

Let `u = 2x` and `v = cos 2x`

So we can write `y = v^3` and `v = cos\ u`

`(dy)/(dx)=(dy)/(dv)(dv)/(du)(du)/(dx)`

`=3v^2(-sin u)(2)`

`=3(cos^2 2x)(-sin 2x)(2)`

`=-6\ cos^2 2x\ sin 2x`

In the final term, put u = 2x³.

We have:

`y=3 sin 4x+5 cos 2x^3`

`(dy)/(dx)=(3)(cos 4x)(4)+` `(5)(-sin 2x^3)(6x^2)`

`=12 cos 4x-30x^2 sin 2x^3`

Put u = 6x² + 5, so y = 4 cos u.

So

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=4[-sin(6x^2+5)][(12x)]`

`=-48x\ sin(6x^2+5)`

Put u = 2x⁴ + 1 and v = sin u

So y = 3v³

`(dy)/(dx)=(dy)/(dv)(dv)/(du)(du)/(dx)`

`=[9v^2][cos u][8x^3]`

`=[9\ sin^2u][cos(2x^4+1)][8x^3]`

`=72x^3sin^2(2x^4+1)cos(2x^4+1)`

Put u = x − cos²x and then y = u⁴.

Now

`(du)/(dx)=1-2\ cos x(-sin x)`

`=1+2\ sin x\ cos x`

and

`(dy)/(du)=4u^3`

So we have:

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=4u^3(du)/(dx)`

`=4[x-cos^2x]^3[1+` `{:2 sin x cos x]`

Put u = 2x + 3 and v = sin 4x

Now

`(dv)/(dx)=4\ cos 4x`

So using the quotient rule, we have:

`(dy)/(dx) =(v(du)/(dx)-u(dv)/(dx))/v^2`

`=((sin 4x)(2)-(2x+3)(4\ cos 4x))/(sin^2 4x)`

`=(2\ sin 4x-4(2x+3)cos 4x)/(sin^2 4x)`

First, we write the right hand side as:

`y = 2x\ sin x + (2 − x^2) cos x`.

We have 2 products. The first term is the product of `(2x)` and `(sin x)`. The second term is the product of `(2-x^2)` and `(cos x)`.

So, using the Product Rule on both terms gives us:

`(dy)/(dx)= (2x) (cos x) + (sin x)(2) +` ` [(2 − x^2) (−sin x) + (cos x)(−2x)]`

`= cos x (2x − 2x) + ` `(sin x)(2 − 2 + x^2)`

`= x^2sin x`

The implicit function:

`x\ cos 2y+sin x\ cos y=1`

We differentiate each term from left to right:

`x(-2\ sin 2y)((dy)/(dx))` `+(cos 2y)(1)` `+sin x(-sin y(dy)/(dx))` `+cos y\ cos x`

`=0`

So

`(-2x\ sin 2y-sin x\ sin y)((dy)/(dx))` `=-cos 2y-cos y\ cos x`

Solving for `dy/dx` gives us:

`(dy)/(dx)=(x(6\ cos 3x)-(2\ sin 3x)(1))/x^2`

`=(6x\ cos 3x-2\ sin 3x)/x^2`

When `x = 0.15` (in radians, of course), this expression (which gives us the slope) equals `-2.65`.

Here is a graph of our situation. The tangent to the curve at the point where `x=0.15` is shown. Its slope is `-2.65`.

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.

This is the product of `x` and `tan x`.

So we have:

`d/(dx)(x\ tan x) =(x)(sec^2x)+(tan x)(1)`

`=x\ sec^2x+tan x`