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# Find derivative of -8e^(xsinx)?

Please help, thank you....this is way beyond what my professor taught us. Thanks in advance.

### 2 Answers

- gudspelingLv 71 decade agoFavorite Answer
y = -8e^(x.sin(x))

u = x.sin(x)

du/dx = sin(x) + x.cos(x)

y = -8e^u

dy/du = -8e^u = -8e^(x.sin(x))

dy/dx = (dy/du)(du/dx) = -8(sin(x)+x.cos(x))e^(x.sin(x))

- StephenLv 51 decade ago
d/dx(-8 e^(x sin(x))) = -8 e^(x sin(x)) (sin(x)+x cos(x))

Possible derivation:

d/dx(-8 e^(x sin(x)))

| Factor out constants:

= | -8 (d/dx(e^(x sin(x))))

| Use the chain rule, d/dx(e^(x sin(x))) = ( de^u)/( du) ( du)/( dx), where u = x sin(x) and ( de^u)/( du) = e^u:

= | -8 e^(x sin(x)) (d/dx(x sin(x)))

| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = sin(x):

= | -8 e^(x sin(x)) (x (d/dx(sin(x)))+sin(x) (d/dx(x)))

| The derivative of sin(x) is cos(x):

= | -8 e^(x sin(x)) sin(x) (d/dx(x))-8 x e^(x sin(x)) cos(x)

| The derivative of x is 1:

= | -8 e^(x sin(x)) sin(x)-8 x e^(x sin(x)) cos(x)

or you may have meant

d/dx(-8 e^x sin(x)) = -8 e^x sin(x)-8 e^x cos(x)