Question #285884

Find the derivatives of all orders of the functions in Exercises 29 - 32 . 29. y = (x ^ 4)/2 - 3/2 * x ^ 2 - x 30. y = (x ^ 5)/120 30. y = (x - 1)(x + 2) * (x + 3) 32. y = (4x ^ 2 + 3)(2 - x) * x


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Expert's answer
2022-01-10T14:51:56-0500

29).y=x423x22xy=4x326x21y=2x33x1.30).y=x5120y=5x4120y=x42431).y=(x1)(x+2)(x+3)we use product rules whereu=(x1),dudx=1,v=(x+2),dvdx=1,w=(x+3),dwdx=1uvdwdx+vwdudx+uwdvdxy=(x1)(x+2)+(x+2)(x+3)+(x1)(x+3)32).y=(4x2+3)(2x)xwe use product rules whereu=(4x2+3),dudx=8x,v=(2x),dvdx=1,w=x,dwdx=1uvdwdx+vwdudx+uwdvdxy=(4x2+3)(2x)+8x2(2x)x(4x2+3)29).\\y=\frac{x^4}{2}-\frac{3x^2}{2}-x\\ y'=\frac{4x^3}{2}-\frac{6x}{2}-1\\ y'=2x^3-3x-1.\\ 30).\\ y=\frac{x^5}{120}\\ y'=\frac{5x^4}{120}\\ y'=\frac{x^4}{24}\\ 31).y=(x-1)(x+2)(x+3)\\ \text{we use product rules where}\\ u=(x-1), \frac{du}{dx}=1,v=(x+2),\frac{dv}{dx}=1,w=(x+3),\frac{dw}{dx}=1\\ uv\frac{dw}{dx}+vw\frac{du}{dx}+uw\frac{dv}{dx}\\ y'=(x-1)(x+2)+(x+2)(x+3)+(x-1)(x+3)\\ 32).\\ y=(4x^2+3)(2-x)x\\ \text{we use product rules where}\\ u=(4x^2+3), \frac{du}{dx}=8x,v=(2-x),\frac{dv}{dx}=-1,w=x,\frac{dw}{dx}=1\\ uv\frac{dw}{dx}+vw\frac{du}{dx}+uw\frac{dv}{dx}\\ y'=(4x^2+3)(2-x)+8x^2(2-x)-x(4x^2+3)\\


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