you 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!
okay i need to find the derivative of
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!
4 answers
Hey, thank you so much! you did help!
the second is x squared.. sorry for no being clear
the second is x squared.. sorry for no being clear
π 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)
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)
PS - 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
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
1 answer