Question

Find the derivative using the quotient rule. See attachment for equation.

Answer 1

Step 1:

Write the function

`\text{y = }\frac{4x^3-3x^2}{4x^5\text{ - 4}}`

Step 2:

Apply the quotient rule below to find the derivative of the function.

`\begin{gathered} \text{If y = }(u)/(v) \\ \frac{d\text{ y}}{d\text{ x}}\text{ = }\frac{v(du)/(dx)\text{ -u}\frac{d\text{ v}}{d\text{ x}}}{v^2} \\ u=4x^3-3x^2 \\ v=4x^5-4_{}_{} \end{gathered}`

Step 3:

`\begin{gathered} u=4x^3-3x^2 \\ \frac{d\text{ u}}{d\text{ x}}=12x^2\text{ - 6x} \\ v=4x^5\text{ - 4} \\ \frac{d\text{ v}}{d\text{ x}}=20x^4 \end{gathered}`

Step 4:

Substitute

`\begin{gathered} \frac{d\text{ y}}{d\text{ x}}\text{ = }\frac{(4x^5-4)(12x^2-6x)-20x^4(4x^3-3x^2)^{}}{(4x^5-4)^2} \\ =\text{ }(48x^7-24x^6-48x^2+24x-80x^7+60x^6)/((4x^5-4)^2) \\ =\text{ }\frac{-32x^7+36x^6-48x^2+24x^{}}{(4x^5-4)^2} \\ =\text{ }\frac{4x(-8x^6+9x^5\text{ -12x +6)}}{16(x^5-1)^2} \\ Final\text{ answer} \\ =\text{ }\frac{x(-8x^6+9x^5\text{ - 12x + 6)}}{4(x^5-1)^2} \end{gathered}`