# Definition of the Derivative, Product and Quotient Rules

1. Differentiate from first principles( for x radians):

a) sin x

b) cos x

2. Products and quotients

For a function, f(x), which can be expressed as a product or quotient of other functions, u(x) and v(x), there exist

a) the product rule, f(x) = u(x) ? v(x),

df/dx = u ? dv/dx + v ? du/dx, and

b) the quotient rule, f(x) = u(x)/v(x),

df/dx =( v? du/dx - u? dv/dx) /v²

Prove these rules form first principles. It can be assumed that u(x) and v(x) are both differentiable.

3. Negative powers of x

Prove form first principles that for f(x) = xⁿ, where n < 0, df/dx = nx ⁿ ‾ ¹

The result for f(x) = xⁿ, where n > 0, can be assumed; as can the sum, difference, product and quotient rules ( if / as appropriate).

4. For f(x) = xⁿ, where n > 0, df/dx = nx. For f(x) =x, where n = 0, f(x) is a constant.

Which function, f(x), has the derivative df/dx = x‾ ¹ = 1/x ?

Prove this from first principles - from the definition of the derivative.

#### Solution Preview

NOTE: In the following, all limits in symbol of "lim " mean taking limit when h goes to 0.

1. Differentiate from first principles( for x radians):

a) sin x

Proof.

[sinx]'=lim [sin(x+h)-sinx]/h

=lim [2cos(x+h/2)sin(h/2)]/h

=cosx*lim sin(h/2)/(h/2)

=cosx

since lim sin(h/2)/(h/2)=1

b) cos x

Proof.

[cosx]'=lim [cos(x+h)-cosx]/h

=lim [-2sin(x+h/2)sin(h/2)]/h

=-sinx*lim sin(h/2)/(h/2)

=-sinx

since lim sin(h/2)/(h/2)=1

2. Products and quotients

For a function, f(x), which can be expressed as a product or quotient of other functions, u(x) and v(x), there exist

a) the product rule, f(x) = u(x) ? v(x),

df/dx = u ? dv/dx + v ? du/dx,

Proof. If f(x) = u(x) ? v(x), then

df/dx=lim [f(x+h)-f(x)]/h

=lim [u(x+h)v(x+h)-u(x)v(x)]/h

=lim [u(x+h)v(x+h)-u(x)v(x+h)+u(x)v(x+h)-u(x)v(x)]/h

...

#### Solution Summary

Derivatives are found and the product and quotient rules are proven from first principles. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.