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) = 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.
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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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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
...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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