Question

What is the derivative of y=xsinx+cosx? Is it simply y'=1cosx-sinx because derivative of x=1, derivative of sinx=cosx and derivative of cosx=-sinx?

Answer 1

\[y' = [x(\cos x) + \sin x] - \sin x = x(\cos x)\]You just use the product rule on that first term to get the value within the brackets.

Answer 2

If y = uv then y' = u'v + uv'

Answer 3

only if you are taking the derivative "with respect to x" is x' = 1

Answer 4

All good now, @schi1 ?

Answer 5

still not sure where the extra -sinx came from

Answer 6

the derivative of the cos is the negative sine

Answer 7

the derivative of sinx should equal cosx
and the derivative of cosx should equal -sinx shouldn't it?

Answer 8

so the derivative of y=xsinx+cosx should just be cosx-sinx isn't it?

Answer 9

y = xf+g

y' = (xf)' + (g)'

y' = x'f + xf' + g' , if this is wrt.x, x'=1

y' = f + xf' + g'

Answer 10

Yes, that's exactly right, @schi . In that original function, you have 2 terms. The first one shows the derivative in the brackets. You still have to do the derivative of the second term.

Answer 11

ohh ok gotcha I missed a rule

Answer 12

In your problem, the way it is written, that "x" gets multiplied only to sin x indicating the derivation that @amistre64 so very well pointed out. If that is how the problem is indeed intended. If, however you want "x" multiplied to the sum of sinx and cos x, you have to put that in parentheses.

Answer 13

no x only multiplies sinx

Answer 14

ok, that's good. Then that first post of mine gives the actual derivative.

Answer 15

So, to sum up, you get the derivative of x sin x -> x cos x + sin x

And then you get the derivative of cos x -> - sin x

And then you add those. The sin x and the "- sin x" will cancel.

Answer 16

You should be all set now, @schi1 . Are you ok with the answer now?

Answer 17

yes thank you