Question

Differentiate sinxcosx

Answer 1

product rule

u =sinx du = cosx
v = cosx dv = - sinx

dy/dx = -sin^2x + cos^2x

or cos^2(x) - sin^2(x)

= cos2x (if necessary)

Answer 2

sinxcosx = 1/2 (sin2x)
d(sinxcosx) / dx = d/dx [1/2 (sin2x)]
= 1/2 d/dx (sin2x)
= 1/2 (2) (cos 2x)
= cos2x

Answer 3

I understand how you differentiated u and v, but how did you get the -sin to have a power to 2, and same with cos?

Answer 4

y=(left)(right)
y'= (copy left)(differentiate right) + (copy right)(differentiate left)

In your case,

\[y=sinxcosx\]
\[y'=sinx(-sinx) + cosx(cosx)\]
\[y'=\cos^2 x -\sin^2 x\]

Answer 5

the derivative of cos(x) is - sin(x) one of those basic facts when differentiating trig function

Answer 6

Noneed to memorize product rule formula

Answer 7

i think the better way is to use the double angle formula first.
ie : sinxcosx = 1/2 (sin2x)
that helps reduce steps later on when you diff. the expression

Answer 8

I know that cosx = -sin x when differentiating.. but @campbell_st , how does that change into ^2 once using dx/dy?

@.Sam. Thanks, but how did you get y' ?

Answer 9

becuase of the product rule says u x v' so sin(x) x (-sin(x)) = -sin^2(x)

Answer 10

Ah Ok. That makes more sense. Thanks!!

Answer 11

no problem