How do I differentiate a three-term product using the product rule?

[McKendrigo]

Homework Statement

Using the product rule, differentiate the following function:

Homework Equations

y = e^tsintcost

The Attempt at a Solution

The three term product rule says:

d/dx (uvw) = u'vw + uv'w + uvw'

I find u = e^t, u' = e^t, v = sint, v' = cost, w = cost and w' = -sint

Thus, dy/dx = e^tsintcost + e^tcos²t - e^tsin²t

= e^t(sintcost + cos²t -sin²t)

and since cos²t + sin²t = 1 we can finally re-write this as:

dy/dx = = e^t(sintcost - 1)

However my textbook answer states that:
dy/dx = = e^t(2cos²t + sintcost - 1)

I'm really not sure where this extra 2cos²t term comes from. Any help?

Thanks in advance!

 

[Mentallic]

Homework Statement

Using the product rule, differentiate the following function:

Homework Equations

y = e^tsintcost

The Attempt at a Solution

The three term product rule says:

d/dx (uvw) = u'vw + uv'w + uvw'

I find u = e^t, u' = e^t, v = sint, v' = cost, w = cost and w' = -sint

Thus, dy/dx = e^tsintcost + e^tcos²t - e^tsin²t

= e^t(sintcost + cos²t -sin²t)

and since cos²t + sin²t = 1 we can finally re-write this as:

dy/dx = = e^t(sintcost - 1)

However my textbook answer states that:
dy/dx = = e^t(2cos²t + sintcost - 1)

I'm really not sure where this extra 2cos²t term comes from. Any help?

Thanks in advance!

Homework Statement

Homework Equations

The Attempt at a Solution

Your mistake was in the substitution you made, if $$sin^2x+cos^2x=1$$ then $$-(sin^2x+cos^2x)=-1$$ and that isn't what you had in the derivative equation.

 

[McKendrigo]

Arrrrrrgh! Thanks for pointing that out!

Use the double angle formula:

cos²t - sin²t = 2cos²t -1

instead and it all works out fine.

Thanks again for your help.

 

[Mentallic]

Yep Equivalently just rearrange the basic equation $$sin^2x+cos^2x=1$$ to $$sin^2x=1-cos^2x$$ and substitute in, but you probably already know that!