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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:

Differentiation Interactive Applet - trigonometric functions.

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:


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


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


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.

Graph y = cos(x^2+3)

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!

Graph y = cos(x^2+3)

Example 2

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


Let u = 3x4 and so `y = cos u`.



`=-sin u(du)/(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`


`=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`


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


Put u = 6x2 + 5, so y = 4 cos u.




`=-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


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

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


3. Differentiate y = (x − cos2x)4.


Put u = x − cos2x and then y = u4.


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

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



So we have:



`=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


`(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`



`(-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`.

Graph of tangent to a curve

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)`


`=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:


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:


`=[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`


` ="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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See also: Derivative of square root of sine x by first principles.

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