What is the derivative of xsin⁡x

rheisf

rheisf

Answered

2021-12-29

What is the derivative of xsinx

Answer & Explanation

Bernard Lacey

Bernard Lacey

Expert

2021-12-30Added 30 answers

dydx=xcosx+sinx
Explanation:
We have:
y=xsinx
Which is the product of two functions, and so we apply the Product Rule for Differentiation:
ddx(uv)=udvdx+dudxv,  or  (uv)=(du)v+u(dv)
I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".
So with y=xsinx
{Letu=xdudx=1And v=sinxdvdx=cosx
Then:
ddx(uv)=udvdx+dudxv
Gives us:
ddx(xsinx)=(x)(cosx)+(1)(sinx)
dydx=xcosx+sinx
If you are new to Calculus then explicitly substituting u and v can be quite helpful, but with practice these steps can be omitted, and the product rule can be applied as we write out the solution.
Maricela Alarcon

Maricela Alarcon

Expert

2021-12-31Added 28 answers

This is a function which is in the form,
y=f(x)g(x)
Its
Vasquez

Vasquez

Expert

2022-01-08Added 457 answers

Finally found the answer to this difficult question, I'm happy

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