# 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`

In words, we would say:

The derivative of sin

xis cosx,

The derivative of cosxis −sinx(note the negative sign!) and

The derivative of tanxis sec^{2}x.

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)`.

### Example 2

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

### Example 3

Differentiate `y = cos^3 2x`

### Example 4

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

## Exercises

1. Differentiate *y* = 4 cos (6*x*^{2} + 5).

2. Find the derivative of *y* = 3 sin^{3} (2*x*^{4} + 1).

3. Differentiate *y* = (*x* − cos^{2}*x*)^{4}.

4. Find the derivative of:

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

5. Differentiate *y* = 2*x* sin *x* + 2 cos *x* − *x*^{2}cos *x*.

6. Find the derivative of the implicit function

xcos 2y+ sinxcosy= 1.

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

`y=(2\ sin\ 3x)/x`

where `x = 0.15`

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)`

9. Show that *y* = cos^{3}*x* tan *x* satisfies

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

10. Find the derivative of *y *=* x* tan *x*

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

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