What are derivatives (in calculus)?
Topic: Problem solving use objects and reasoning
July 18, 2019 / By Roni Question:
I've been working on an assignment dealing with the relationships between derivatives and various problems. (For example, I have to explain how derivatives are used to solve maximum and minimum problems right now) However, i can't seem to find a simple definition or example of what derivatives are. In some places, it says it has to do with instantaneous velocity, and in other places, it seems to have to do with a function.
Although this is a graduate level calculus class, I have never taken calculus before, so all of this is new to me. Any help or suggestions would be appreciated greatly!!
Thanks Denis! I can't believe no one else could explain it like that.
Best Answers: What are derivatives (in calculus)?
Monat | 7 days ago
when youre talking about a derivative, youre taking about the instantaneous rate of change is how fast something is changing at that instant. so when youre talking about cost, the derivative is the refers to how fast the price of something is changing at a particular instant. or in your case, when your talking about velocity, the instantaneous velocity refers to how fast the velocity is changing at a certain point on the map. u can use the derivative and plug in a value of x or t or whichever input value into the derivative to find the slope of the tangent line at that instant. you can also have a case where u have to take the dervitive with respect to something else. for example, let S = position and let t = time. dS/dt would refer to the instantaneous rate of change of position with respect to time, which means how fast an object is moving (changing position) at a given time. note that there is also something called the average rate of change, which is often referred to the slope of the secant line. this is not the same thing as the derivative. the slope of the secant line is found by taking two points at a certain function and slope between the two points. the slope of the tangent line, or instantaneous rate of change is when u inch up to a certain point on the function and find the slope at that particular instant, and that is the essence of the derivative.
about maximums and minimums, derivitives are useful for those. when a function has a positive derivative that means that function is increasing at that particular instant. lets say f'(2) = 5. this means at that instant when x is equal to 2, the slope of the line tangent to f(x) at that instant is postive. so we know that the function is increasing. similarly if the derivative was negative at a certain instant, we know that the slope is negative at that particular instant and that the function is decreasing. for a there to be a min the derivitive must go from negative to 0 to positive. this means that the function was had a negative slope on an interval of values (which means it was decreasing), became stationary at a certain instant (when the derivative = 0), and had a positive slope on the next interval (which implies that it was increasing. when a function goes from decreasing to stationary to increasing, that implies that there is a minimum at that stationary point. draw it out for yourself. u would use the same reasoning to deduce the maximum
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Originally Answered: Calculus derivatives homework help?
For the first one, I'm not sure if you know implicit and explicit differentiation. Implicit is just finding the derivative as the function is. Explicit is changing around the function into something like y = something.
I'll do explicit differentiation.
A. xy + 2x + 3y = 1
B. xy + 3y = 1 - 2x
C. y(x + 3) = 1 - 2x
D. y = (1 - 2x) / (x + 3)
E. y' = [ (x + 3)(-2) - (1 - 2x) (1) ] / (x + 3)²
F. y' = [ (-2x - 6) - (1 - 2x) ] / (x + 3)²
G. y' = -5 / (x + 3)²
B. Subtract 2x to the other side.
C. Factor out a y from the left side.
D. Divide by (x + 3) to get y alone.
E. Start differentiating by using the quotient rule.
F. Distribute and simplify.
Here's the implicit way.
A. xy + 2x + 3y = 1
B. x(dy/dx) + y(1) + 2 + 3(dy/dx) = 0
C. x(dy/dx) + 3(dy/dx) = -y - 2
D. (dy/dx) (x + 3) = -y - 2
E. (dy/dx) = (-y - 2) / (x + 3)
B. Differentiate each term. The derivative of y is (dy/dx). Remember your product rules.
C. Subtract y and 2 to isolate (dy/dx)
D. Factor our a (x + 3).
E. Divide by (x + 3) and you're done.
Note: (dy/dx) means derivative.
The second one I'm not as sure with.
A. y = zcos^-1(z) - (1 - z²)^(1/2)
B. y' = [ z(-1 / √(1 - z²) ) + cos^-1(z) (1) ] - [ (1/2)(1 - z²)^(-1/2) (-2z) ]
C. y' = [ (-z / √(1 - z²) ) + cos^-1(z) ] - [ -2z / ( 2 (1 - z²)^(1/2) ) ]
D. y' = [[-z + (cos^-1(z) √(1 - z²)) ] / √(1 - z²) ] + [ -2z / 2√(1 - z²) ]
E. y' = [[-z + (cos^-1(z) √(1 - z²)) ] / √(1 - z²) ] + [ -z / √(1 - z²) ]
F. y' = [-z + (cos^-1(z) √(1 - z²)) - z] / √(1 - z²)
G.y' = [-2z + (cos^-1(z) √(1 - z²))] / √(1 - z²)
B. Product rules and power rules.
C. Distributing z and -2z.
D. Finding a common denominator for the first part.
E. Canceling out a 2 from the second part of the equation.
F. Combining like terms since everything has a common denominator. Note: ^(1/2) is the same as √.
G. Answer, but I'm not completely sure..
I think if derivatives as the instantaneous change in y with respect to x (assuming x and y are labels for the 2 dimensions). So for a curve, dy/dx is the slope of the curve for any value of x. If a curve has a point where the slope is 0 (for a quadratic, e.g.), it must be either a max or a min. For higher power relationships, this is not necessarily so. If x and y represent distance (s) and time (t), then ds/dt would represent the instantaneous velocity.
Hope this helps.
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The derivative is the slope of the tangent line to the curve at any point.
If your function is f(x)=x^2, then the derivative is f'(x)=2x.
The slope of f(x) at x=1 would be f'(1)=2.
You could then find the equation of the tangent line since you know the slope and a point (1,1) using the point-slope form of a line.
That's the simplest defintion of what the derivative is.
You can then find relative extrema (minima, maxima) when the derivative goes to zero. Again, back at f(x)=x^2. It has an extrema at x=0, because f'(x)=2x=0, when x=0. Then the slope of the tangent line to the curve is 0, in other words, the tangent line is horizontal as you would expect at a minimum or maximum.
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The derivative, a brance of mathematics in Calculus, is defined as the
measure of how a function changes in output as its input changes.
For example: the derivative of the position of an object (or
distance) with the respect to a time ( a given reference point )
is the instananeous velocity or respectively its instantaneous speed.
The process of finding a derivative is called differentiation.
For additional information, please see also:
"May the Lord Jesus Christ continue to richly bless you in all
of your endeavors."
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It may help to do a u substitution. y = (1+1/x)^3 Let u = 1 + 1/x Then y' = f'(u)*u' y = u^3 y' = 3*u^2 * u' y' = 3* (1 + 1/x)^2 * (-1/x^2) For y'' you will also need the product rule. y'' = 3*(1+1/x)^2 * (2/x^3) + (-1/x^2)*derivative ( 3*(1+1/x)^2) Using u substitution again: -------------A------- y'' = [6*(1+1/x)^2]/x^3 -1/x^2 * 6u*u' = A -1/x^2* 6*(1+1/x)*-1/x^2 = A + [6*(1+1/x)] / x^4
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Originally Answered: Taking derivatives (Calculus)?
f(x) = (x^2 - 5x + 2)(x - 2/x)
The above can be rewritten as
f(x) = (x^2 - 5x + 2)(x - 2x^-1)
and then simply apply the product rule.
Differentiating the above,
f' (x) = (x^2 - 5x + 2)(1 + 2/x^2) + (x - 2x^-1)(2x - 5)
f' (x) = (x^2 - 5x + 2)(1 + 2/x^2) + (x - 2/x)(2x - 5)
I trust that you can proceed on your own to simplify the above.
Hope this helps.