Two Derivatives.

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Originally posted by PrudensOptimus Find f'(x) and f'(-1).f(x) = (((1/x^2) - 2)/(1/x^2))2f'(x) = d/dx ((1/x^2) - 2)/(1/x^2) * (2 * ((1/x^2) - 2)/(1/x^2)) = [2((1/x^2) - 2)/(1/x^2)] * [((1/x^2)(-2/x^3) - ((1/x^2)-2)(-2/x^3))/((1/x^2)^2)] = ... * [ -4/x^7
  • #1
PrudensOptimus
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Find the derivative for the f(x).
f(x) = pi*x + 1/(cos^2(Pi*x))
And find f'(1/4)

Here's what i got.
f'(x) = pi * d/dx (x) + d/dx (cos^2 pi*x)-1
= pi + [ (- (d/dx cos^2 (pi*x))/(cos^4(pi*x)) ]
= pi + [ (- (d/dx cos(pi*x) * (2 * cos(pi * x))/(cos^4(pi*x)) ]
= pi + [ (pi * sin x * 2cos(pi * x)) / ... ]
= pi + [ (2 * pi * sin x) / cos^3 (pi * x) ]

Thus f'(1/4) = 7.5251
 
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  • #2
#2 question:

Find f'(x) and f'(-1).

f(x) = (((1/x^2) - 2)/(1/x^2))2


f'(x) = d/dx ((1/x^2) - 2)/(1/x^2) * (2 * ((1/x^2) - 2)/(1/x^2))
= [2((1/x^2) - 2)/(1/x^2)] * [((1/x^2)(-2/x^3) - ((1/x^2)-2)(-2/x^3))/((1/x^2)^2)]
= ... * [ -4/x^7 ]
= -8(1 - 2x^2)/x^7

Thus, f'(-1) = -8.

Please correct my mistakes. Thanks.
 
  • #3
Originally posted by PrudensOptimus
Find the derivative for the f(x).
f(x) = pi*x + 1/(cos^2(Pi*x))
And find f'(1/4)
.
.
.
= pi + [ (2 * pi * sin x) / cos^3 (pi * x) ]

Thus f'(1/4) = 7.5251

I don't have a calculator handy, but your derivative is correct.

Find f'(x) and f'(-1).

f(x) = (((1/x^2) - 2)/(1/x^2))2
.
.
.
= -8(1 - 2x^2)/x^7

Thus, f'(-1) = -8.

For the derivative, I get

f'(x)=-8x+16x3

I rewrote the function as follows:

f(x)=(1/x2-2)2x4

To get the last factor, I noted that 1/(1/x2) is simply x2. Squaring that gives x4.

edit: It would be best to expand the function before differentiating, as follows:

f(x)=(1/x4-4/x2+4)x4

Now distribute the x4

f(x)=(1-4x2+4x4)

Doesn't look as nasty now, does it? :smile:
 
Last edited:
  • #4
You sure you got your #2 derivative right?

It took me two times to re-derive it and everytime I get the same answer.
 
  • #5
Originally posted by PrudensOptimus
Here's what i got.
f'(x) = pi * d/dx (x) + d/dx (cos^2 pi*x)-1
= pi + [ (- (d/dx cos^2 (pi*x))/(cos^4(pi*x)) ]
= pi + [ (- (d/dx cos(pi*x) * (2 * cos(pi * x))/(cos^4(pi*x)) ]
= pi + [ (pi * sin x * 2cos(pi * x)) / ... ]
= pi + [ (2 * pi * sin x) / cos^3 (pi * x) ]

Thus f'(1/4) = 7.5251

Should be this on the fourth line:
= π + π sin(π x) 2cos(π x)/cos4(π x)
i.e. Missed a π inside the sin(x).

And the final answer:
f'(x) = π + 2π sec2(π x)tan(πx)
f'(1/4) ≈ 15.708

For your second question, you surprisingly did get the correct value for f'(-1), but the function you found is in general wrong:
f(x) = [(1/x2-2) / (1/x2)]2
f(x) = [(1/x2-2) *x2]2
f(x) = (1/x2-2)2*x22
f(x) = (1/x4-4/x2+4)x4
f(x) = 1-4x2+4x4
(as Tom derived)

From this it far easier to find f'(x):
f'(x)= -8x+16x3
f'(x) = 8x(2x2-1)
f'(-1) = -8
This is similar to be not the same as the function you derived.
 

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function with respect to a given variable. It is essentially the slope of the tangent line at a specific point on a graph.

2. What is the purpose of finding derivatives?

The main purpose of finding derivatives is to analyze and understand the behavior of a function. It can also be used to solve optimization problems, such as finding the maximum or minimum value of a function.

3. What is the difference between a first derivative and a second derivative?

A first derivative represents the rate of change of a function, whereas a second derivative represents the rate of change of the first derivative. In other words, a second derivative measures the rate of change of slope.

4. How is the derivative calculated?

The derivative of a function is calculated using the rules of differentiation, such as the power rule, product rule, and chain rule. These rules involve manipulating the original function to find the rate of change at a specific point.

5. What are some real-life applications of derivatives?

Derivatives have many real-life applications, such as in physics to calculate velocity and acceleration, in economics to analyze the rate of change of supply and demand, and in engineering to design and optimize structures and systems. It is also used in fields like medicine, finance, and computer science.

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