Tuesday

September 16, 2014

September 16, 2014

Posted by **tim** on Tuesday, November 13, 2012 at 11:16pm.

f(x)=πsin2x

and

f(x)=cosπx^2

----------------

π is pi

and if you could show steps it would be GREATLY appreciated. thank you so much!

- calculus -
**bbbbb**, Tuesday, November 13, 2012 at 11:43pmyou need to use the product rule so the first one(which is pi) times the derivative of the second one(which is sin2x) plus the second one times the derivative of the first one.

Here is how it should look:

(pi*cos2x)+(sin2x*1)The one is because pi is a constant so its derivative is just one.

In the second problem I do not understand what is squared...the x or pi times x. Sorry

I hope I helped you one the first one!

- calculus -
**tim**, Wednesday, November 14, 2012 at 12:37amHey, thank you so much! you did help!

the second is x squared.. sorry for no being clear

- calculus -
**Steve**, Wednesday, November 14, 2012 at 10:38amπ is just a constant, so

d/dx(πx) = π

d/dx(πsin2x) = π(2cos2x) = 2πcos2x

the second one uses the chain rule:

if u is a function of x, and f = cos(u), f' = -sin(u) u'

u = πx^2, so u' = 2πx

d/dx cos(πx^2) = -2πx sin(πx^2)

- calculus -
**Steve**, Wednesday, November 14, 2012 at 10:40amPS - derivative of a constant is 0, not 1.

Reason 1: derivative is rate of change. constants do not change.

Reason 2: since 1 = x^0, a constant k = kx^0

the derivative is thus 0*k*x^(-1) = 0

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