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) ← chain rule
and the standard derivatives
[tex]\frac{d}{dx}[/tex] ([tex]log_{a}[/tex] x ) = [tex]\frac{1}{xlna}[/tex] , [tex]\frac{d}{dx}[/tex](lnx) = [tex]\frac{1}{x}[/tex]
(a)
Given
y = [tex]log_{a}[/tex][tex]\sqrt{(1+x)}[/tex]
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{lna\sqrt{(1+x)} }[/tex] × [tex]\frac{d}{dx}[/tex] ([tex](1+x)^{\frac{1}{2} }[/tex]
= [tex]\frac{1}{lna\sqrt{(1+x)} }[/tex] × [tex]\frac{1}{2}[/tex] [tex](1+x)^{-\frac{1}{2} }[/tex] × [tex]\frac{d}{dx}[/tex] (1 + x)
= [tex]\frac{1}{lna\sqrt{(1+x)} }[/tex] × [tex]\frac{1}{2\sqrt{(1+x)} }[/tex] × 1
= [tex]\frac{1}{2lna(1+x)}[/tex]
= [tex]\frac{1}{(1+x)lna^2}[/tex]
(b)
Given
y = ln sinx
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{sinx}[/tex] × [tex]\frac{d}{dx}[/tex](sinx)
= [tex]\frac{1}{sinx}[/tex] × cosx
= [tex]\frac{cosx}{sinx}[/tex]
= cotx
