Answer:
see explanation
Step-by-step explanation:
Using the chain rule
Given
y = f(g(x)), then
[tex]\frac{dy}{dx}[/tex] = f'(g(x)) × g'(x)
(a)
y = [tex]\frac{2}{(3x^2-x+7)^3}[/tex] = 2 ([tex]3x^2-x+7)^{-3}[/tex]
[tex]\frac{dy}{dx}[/tex] = 2. - 3 [tex](3x^2-x+7)^{-4}[/tex] × [tex]\frac{d}{dx}[/tex] (3x² - x + 7)
= - 6 [tex](3x^2-x+7)^{-4}[/tex] × (6x - 1)
= [tex]\frac{-6(6x-1)}{(3x^2-x+7)^4}[/tex]
---------------------------------------------------
(b)
y = [tex](4x-10)^{10}[/tex]
[tex]\frac{dy}{dx}[/tex] = 10[tex](4x-10)^{9}[/tex] × [tex]\frac{d}{dx}[/tex](4x - 10)
= 10[tex](4x-10)^{9}[/tex] × 4
= 40[tex](4x-10)^{9}[/tex]
