Question

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

Answer 1

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.

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

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