Answer to Question #318544 in Calculus for Zane

Question #318544

Find the derivative using chain rule-d/dx: 


5.y = (4x - 9) * (3x - 4) ^ 2


6.y = x/(x ^ 2 - 2) ^ 2)


7.) y =(x²-7)²(2x - 5)³


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


9. y = sqrt(x - 2) ^ 2


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


1
Expert's answer
2022-03-31T13:41:26-0400

5.


"\\frac{d}{dx}\\left(\\left(4x-9\\right)\\left(3x-4\\right)^2\\right)=\\frac{d}{dx}\\left(4x-9\\right)\\left(3x-4\\right)^2+\\frac{d}{dx}\\left(\\left(3x-4\\right)^2\\right)\\left(4x-9\\right)="


"=4\\left(3x-4\\right)^2+6\\left(3x-4\\right)\\left(4x-9\\right)=108x^2-354x+280"


6.


"\\frac{d}{dx}\\left(\\frac{x}{\\left(x^2-2\\right)^2}\\right)=\\frac{\\frac{d}{dx}\\left(x\\right)\\left(x^2-2\\right)^2-\\frac{d}{dx}\\left(\\left(x^2-2\\right)^2\\right)x}{\\left(\\left(x^2-2\\right)^2\\right)^2}=\\frac{1\\cdot \\left(x^2-2\\right)^2-4x\\left(x^2-2\\right)x}{\\left(\\left(x^2-2\\right)^2\\right)^2}=\\frac{-3x^2-2}{\\left(x^2-2\\right)^3}"


7.


"\\frac{d}{dx}\\left(\\left(x^2-7\\right)^2\\left(2x-5\\right)^3\\right)=\\frac{d}{dx}\\left(\\left(x^2-7\\right)^2\\right)\\left(2x-5\\right)^3+\\frac{d}{dx}\\left(\\left(2x-5\\right)^3\\right)\\left(x^2-7\\right)^2="


"=4x\\left(x^2-7\\right)\\left(2x-5\\right)^3+6\\left(2x-5\\right)^2\\left(x^2-7\\right)^2"


8.


"\\frac{d}{dx}\\left(\\frac{\\left(x^4-2x\\right)^2}{\\left(x-1\\right)^3}\\right)=\\frac{\\frac{d}{dx}\\left(\\left(x^4-2x\\right)^2\\right)\\left(x-1\\right)^3-\\frac{d}{dx}\\left(\\left(x-1\\right)^3\\right)\\left(x^4-2x\\right)^2}{\\left(\\left(x-1\\right)^3\\right)^2}="


"=\\frac{2\\left(x^4-2x\\right)\\left(4x^3-2\\right)\\left(x-1\\right)^3-3\\left(x-1\\right)^2\\left(x^4-2x\\right)^2}{\\left(\\left(x-1\\right)^3\\right)^2}=\\frac{\\left(x^4-2x\\right)\\left(5x^4-8x^3+2x+4\\right)}{\\left(x-1\\right)^4}"


9.


"\\frac{d}{dx}\\left(\\left(\\sqrt{x-2}\\right)^2\\right)=2\\sqrt{x-2}\\frac{d}{dx}\\left(\\sqrt{x-2}\\right)=2\\sqrt{x-2}\\frac{1}{2\\sqrt{x-2}}=1"


10.


"\\frac{d}{dx}\\left(\\left(x^3+x^2\\right)^2\\left(x-1\\right)^3\\right)=\\frac{d}{dx}\\left(\\left(x^3+x^2\\right)^2\\right)\\left(x-1\\right)^3+\\frac{d}{dx}\\left(\\left(x-1\\right)^3\\right)\\left(x^3+x^2\\right)^2="


"=2\\left(x^3+x^2\\right)\\left(3x^2+2x\\right)\\left(x-1\\right)^3+3\\left(x-1\\right)^2\\left(x^3+x^2\\right)^2"


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