Differentiate x.e^x.sinx

Use the Product Rule:
(f * g * h)'(x) = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)

x e^x sin(x)
= 1 * e^x sin(x) + x e^x sin(x) + x e^x cos(x)
= e^x [sin(x) + x sin(x) + x cos(x)]

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Differentiate x eˣ sinx

let u = x eˣ
     v = sinx

Product Rule:
d uv
—— = uv' + vu'
 dx


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d (x eˣ) sinx
—————  =  (x eˣ)•cosx + sinx•[x•eˣ + eˣ•1]
       dx
                     =  xeˣcosx + sinx•[xeˣ + eˣ]

                     =  xeˣcosx + eˣ[x + 1]sinx 

                     =  xeˣcosx + eˣ(x + 1)sinx 

                     =  eˣ[ x cosx + (x + 1)sinx ]            ← ANSWER


Have a good one!
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I assume this means x*e^x*sinx

[f(x)*g(x)*h(x)]'=f'(x)g(x)h(x)+ f(x)g'(x)h(x)+ f(x)g(x)h'(x)

1*e^x*sinx+x*e^x*sinx+x*e^x*cosx

e^x(xsinx+sinx+xcosx)