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# Definition of the Derivative, Product and Quotient Rules

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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) = x&#8319;, where n < 0, df/dx = nx &#8319; &#8254; ¹

The result for f(x) = x&#8319;, where n > 0, can be assumed; as can the sum, difference, product and quotient rules ( if / as appropriate).

4. For f(x) = x&#8319;, 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&#8254; ¹ = 1/x ?

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

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#### 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.

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