derivative of -5e^(xsinx)??

Answer 1

-5(1*sinx+xcosx) e^(xsinx)

Hint : ( e^u ) ' = u' e^u

Answer 2

You have two things going on: the chain rule and the product rule. The chain rule says that if you want to take the derivative of f(g(x)), then that's f'(g(x)g'(x).

Here f(x) = -5e^x, g(x) = x sin(x).

f'(x) is easy since it's an exponential function: f'(x) = -5 e^x

To find g'(x) we have to use the product rule: (a(x) * b(x))' = a'(x) b(x) + a(x) b'(x). Here g(x) = a(x)b(x) where a(x) = x, b(x) = sin(x), so g'(x) is

g'(x) = sin(x) + x cos(x).

then f'(g(x)) is

-5e^(x sin(x)) * (sin(x) + x cos(x))

Answer 3

-5e^(xsinx)

-5 d/dx e^(xsinx)

use the chain rule;
let u = x sinx

then d/du (e^u) = e^u

d/dx (x sinx)
use product rule:
d/dx (x) (sinx) + d/dx (sinx) (x)
1(sinx) +x cosx

so d/dx (xsinx) = sinx + xcosx

-5 * e^(x sinx) * (sinx + x cosx) <== answer

Answer 4

-5e^(xsinx)*(sinx+xcosx)

Chain and product rules

Answer 5

for x sin(x), use the product rule.
u=x v=sin(x)
udv+vdu
-5 [ e^(xsin(x)) (sin(x) + xcos(x))]
-5sin(x)e^(xsin(x)) -5xcos(x) e^(xsin(x))