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# 1. Derivatives of the Sine, Cosine and Tangent Functions

by M. Bourne

It can be shown from first principles that:

(d(sin x))/(dx)=cos x

(d(cos x))/dx=-sin x

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

Explore animations of these functions with their derivatives here:

In words, we would say:

The derivative of sin x is cos x,
The derivative of cos x is −sin x (note the negative sign!) and
The derivative of tan x is sec2x.

Now, if u = f(x) is a function of x, then by using the chain rule, we have:

(d(sin u))/(dx)=cos u(du)/(dx)

(d(cos u))/dx=-sin u(du)/(dx)

(d(tan u))/(dx)=sec^2u(du)/(dx)

### Example 1

Differentiate y = sin(x^2 + 3).

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! ### Example 2

Find the derivative of y = cos 3x^4.

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

### Example 3

Differentiate y = cos^3 2x

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

### Example 4

Find the derivative of y = 3 sin 4x + 5 cos 2x^3.

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

## Exercises

1. Differentiate y = 4 cos (6x2 + 5).

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)

2. Find the derivative of y = 3 sin3 (2x4 + 1).

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)

3. Differentiate y = (x − cos2x)4.

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]

4. Find the derivative of:

y=(2x+3)/(sin 4x)

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)

5. Differentiate y = 2x sin x + 2 cos xx2cos x.

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

6. Find the derivative of the implicit function

x cos 2y + sin x cos y = 1.

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)=(-cos 2y-cos y\ cos x)/(-2x\ sin 2y-sin x\ sin y)

= (cos 2y+cos x\ cos y)/(2x\ sin 2y+sin x\ sin y)

7. Find the slope of the line tangent to the curve of

y=(2 sin 3x)/x

where x = 0.15

(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. 8. The current (in amperes) in an amplifier circuit, as a function of the time t (in seconds) is given by

i = 0.10 cos (120πt + π/6).

Find the expression for the voltage across a 2.0 mH inductor in the circuit, given that

V_L=L(di)/(dt)

 V_L =L(di)/(dt)

=0.002(di)/(dt)

=0.002(0.10)(120pi) xx(-sin(120pit+pi/6))

=-0.024pi\ sin(120pit+pi/6)

9. Show that y = cos3x tan x satisfies

cos x(dy)/(dx)+3y sin x-cos^2x=0

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.

10. Find the derivative of y = x tan x

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

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