# Two Derivatives.

PrudensOptimus
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

## Answers and Replies

PrudensOptimus
#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.

Staff Emeritus
Gold Member
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?

Last edited:
PrudensOptimus
You sure you got your #2 derivative right?

It took me two times to re-derive it and everytime I get the same answer.

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:
= &pi; + &pi; sin(&pi; x) 2cos(&pi; x)/cos4(&pi; x)
i.e. Missed a &pi; inside the sin(x).

And the final answer:
f'(x) = &pi; + 2&pi; sec2(&pi; x)tan(&pi;x)
f'(1/4) &asymp; 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.