Find the derivative y=x^5e^3x-e^-x
T TempTation4u619 New member Joined May 8, 2009 Messages 1 May 8, 2009 #1 Find the derivative y=x^5e^3x-e^-x
D Deleted member 4993 Guest May 8, 2009 #2 Re: QUIZ IN 1 HOUR TempTation4u619 said: Find the derivative y=x^5e^3x-e^-x Click to expand... use addition law ddx[u(x)+v(x)] = dudx +dvdx\displaystyle \frac{d}{dx}[u(x) + v(x)] \, = \, \frac{du}{dx} \, + \frac{dv}{dx}dxd[u(x)+v(x)]=dxdu+dxdv and product law ddx[u(x)⋅v(x)] = v⋅dudx +u⋅dvdx\displaystyle \frac{d}{dx}[u(x) \cdot v(x)] \, = \, v \cdot \frac{du}{dx} \, + u \cdot \frac{dv}{dx}dxd[u(x)⋅v(x)]=v⋅dxdu+u⋅dxdv Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you
Re: QUIZ IN 1 HOUR TempTation4u619 said: Find the derivative y=x^5e^3x-e^-x Click to expand... use addition law ddx[u(x)+v(x)] = dudx +dvdx\displaystyle \frac{d}{dx}[u(x) + v(x)] \, = \, \frac{du}{dx} \, + \frac{dv}{dx}dxd[u(x)+v(x)]=dxdu+dxdv and product law ddx[u(x)⋅v(x)] = v⋅dudx +u⋅dvdx\displaystyle \frac{d}{dx}[u(x) \cdot v(x)] \, = \, v \cdot \frac{du}{dx} \, + u \cdot \frac{dv}{dx}dxd[u(x)⋅v(x)]=v⋅dxdu+u⋅dxdv Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you
