What is the derivative of xsinx?
1 Answer
Sep 15, 2017
dy/dx = xcosx+sinx
Explanation:
We have:
y = xsin x
Which is the product of two functions, and so we apply the Product Rule for Differentiation:
d/dx(uv)=u(dv)/dx+(du)/dxv , 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
{ ("Let", u = x, => (du)/dx = 1), ("And" ,v = sinx, => (dv)/dx = cosx ) :}
Then:
d/dx(uv)=u(dv)/dx + (du)/dxv
Gives us:
d/dx( xsinx) = (x)(cosx)+(1)(sinx)
:. dy/dx = xcosx+sinx
If you are new to Calculus then explicitly substituting