Answer to Question #315751 in Calculus for Zane

Question

Find the derivative of the given function.

1.) y = (x ^ 3 - 3)(x ^ 2 + 4x + 1)

2.) y = (7x + 3)(x ^ 4 - x ^ 3 - 9x)

3.) y = x ^ 2 * (2x ^ 2 + x + 1)

4.) y = (x ^ 2 - 10x + 2)(x ^ 3 - 2x ^ 2 + 1)

5.) y=(x^ 2 -2x-1)(x+)

6.) y= sqrt x( x^ 3 -2x^ 2 +7x-1)

7.) t = x ^ (1/3) * (x ^ 2 - 5x + 2)

8.) y = (5x ^ 2 - 13)(x ^ 3 + 6x + 1)

9.) y=x^ -2 (x^ 2 +7

10.) x ^ - 5 * (x ^ 2 + 10x - 5)

Answer 1

$1. y=(x^3-3)(x^2+4x+1)\\ y'=3x^2(x^2+4x+1)+(x^3-3)(2x+4)=\\=5x^4+16x^3+3x^2-6x-12\\ 2. y=(7x+3)(x^4-x^3-9x)\\ y'=7(x^4-x^3-9x)+(7x+3)(4x^3-3x^2-9)=\\ =35x^4-16x^3-9x^2-126x-27\\ 3. y=x^2(2x^2+x+1)\\ y'=2x(2x^2+x+1)+x^2(4x+1)=\\ =8x^3+3x^2+2x\\ 4. y=(x^2-10x+2)(x^3-2x^2+1)\\ y'=(2x-10)(x^3-2x^2+1)+\\+(x^2-10x+2)(3x^2-4x)=\\ =5x^4-48x^3+66x^2-6x-10\\ 5. y=(x^2-2x-1)(x+1)\\ y'=(2x-2)(x+1)+(x^2-2x-1)=\\ =3x^2-2x-3\\ 6. y=\sqrt{x^3-2x^2+7x-1}\\ y'=\frac{1}{2\sqrt{x^3-2x^2+7x-1}}(3x^2-4x+7)\\ 7. t=x^{\frac{1}{3}}(x^2-5x+2)\\ t'=\frac{1}{3}x^{-\frac{2}{3}}(x^2-5x+2)+x^{\frac{1}{3}}(2x-5)\\ 8. y=(5x^2-13)(x^3+6x+1)\\ y'=10x(x^3+6x+1)+(5x^2-13)(3x^2+6)=\\ =25x^4+51x^2+10x-78\\ 9. y=x^{-2}(x^2+7)\\ y'=-2x^{-3}(x^2+7)+x^{-2}\cdot2x\\10. y=x^{-5}(x^2+10x-5)\\ y'=-5x^{-6}(x^2+10x-5)+x^{-5}(2x+10)$