Question

Who can help me d e f thanks​

Answer 1

d)

`y = (2ax^2 + c)^2 (bx^2 - cx)^(-1)`

Product rule:

`y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^(-1) + (2ax^2+c)^2 \bigg((bx^2-cx)^(-1)\bigg)'`

Chain and power rules:

`y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^(-1) - (2ax^2+c)^2 (bx^2-cx)^(-2) \bigg(bx^2-cx\bigg)'`

Power rule:

`y' = 2(2ax^2+c)(4ax) (bx^2-cx)^(-1) - (2ax^2+c)^2 (bx^2-cx)^(-2) (2bx - c)`

Now simplify.

`y' = (8ax (2ax^2+c))/(bx^2 - cx) - ((2ax^2+c)^2 (2bx-c))/((bx^2-cx)^2)`

`y' = (8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c))/((bx^2-cx)^2)`

e)

`y = (3bx + ac)/(√(ax))`

Quotient rule:

`y' = (\bigg(3bx+ac\bigg)' √(ax) - (3bx+ac) \bigg(√(ax)\bigg)')/(\left(√(ax)\right)^2)`

`y'= (\bigg(3bx+ac\bigg)' √(ax) - (3bx+ac) \bigg(√(ax)\bigg)')/(ax)`

Power rule:

`y' = (3b √(ax) - (3bx+ac) \left(-\frac12 \sqrt a \, x^(-1/2)\right))/(ax)`

Now simplify.

`y' = \frac{3b \sqrt a \, x^(1/2) + \frac{\sqrt a}2 (3bx+ac) x^(-1/2)}{ax}`

`y' = (6bx + 3bx+ac)/(2\sqrt a\, x^(3/2))`

`y' = (9bx+ac)/(2\sqrt a\, x^(3/2))`

f)

`y = \sin^2(ax+b)`

Chain rule:

`y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'`

`y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'`

`y' = 2a \sin(ax+b) \cos(ax+b)`

We can further simplify this to

`y' = a \sin(2(ax+b))`

using the double angle identity for sine.