Differentiation Question?

Question

I know that xe^x differentiates to xe^x +e^x, but I dont know how... Could someone please explain it to me? Thanks

Answer 1

Recall the rule for the derivative of a product:

d/dx(uv) = v*du/dx + u*dv/dx.

(You've been relying on this all along; only most of the time, one of the terms has been a constant and so its derivative was zero.)

If u = x and v = e**x, then
d/dx (x * e**x) = e**x * d/dx(x) + x * d/dx(e**x)
= e**x * 1 + x * e**x
= x * e**x + e**x. Q.E.D.

Answer 2

Since there are 2 "x" terms( x and e^x), you must use the product rule:

d/dx uv = d/dx(u) v + d/dx (v) u

When you differentiate exponential equations, just follow the following formula:

If y = e^ (f(x)),
then dy/dx = d/dx (f(x)) x e^(f(x))

With this,

d/dx xe^x = e^x + xe^x

Answer 3

you have to use the product rule for differentiation.
the product rule states take the derivative of the first term times the second term and add it to the first term times the derivative of the second.
x*e^x
1 * e^x + x *e^x
=e^x + xe^x

Answer 4

xe^x
u have to use the uv rule d/dxof(x)(e^x)+d/dxofe^x(x)
d/dx(xe^x)=(1)e^x+x(e^x)
since d/dxofx=1 and d/dxof e^x=e^x

Answer 5

as per the uv rule of differentiation u*v says
u*derivative of v+v*derivative of u

Answer 6

When do I have to learn that??