Answer:
see explanation
Step-by-step explanation:
Note that [tex]log_{e}[/tex] x = lnx ( natural logarithm)
Using the standard derivatives
[tex]\frac{d}{dx}[/tex] (lnx ) = [tex]\frac{1}{x}[/tex] , [tex]\frac{d}{dx}[/tex] ([tex]log_{a}[/tex] x ) = [tex]\frac{1}{xlna}[/tex]
Using the chain rule
Given y = f(g(x)) , then
[tex]\frac{dy}{dx}[/tex] = f'(g(x))) × g'(x) ← chain rule
(c)
y = [tex]log_{10}[/tex] (x² - 2) then
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{(x^2-2)ln10}[/tex] × [tex]\frac{d}{dx}[/tex] (x² - 2)
= [tex]\frac{1}{(x^2-2)ln10}[/tex] × 2x
= [tex]\frac{2x}{(x^2-2)ln10}[/tex]
(d)
y = ln([tex]\frac{1}{x}[/tex] )
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{\frac{1}{x} }[/tex] × [tex]\frac{d}{dx}[/tex] ([tex]x^{-1}[/tex] )
= x × - [tex]x^{-2}[/tex]
= x × [tex]\frac{-1}{x^2}[/tex]
= - [tex]\frac{1}{x}[/tex]
