Question

Hi. I need help with these questions (see image)Please show workings.​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the chain rule

Given

y = f(g(x)), then

$\frac{dy}{dx}$ = f'(g(x))  × g'(x) ← chain rule

and the standard derivatives

$\frac{d}{dx}$ ($log_{a}$ x ) = $\frac{1}{xlna}$ , $\frac{d}{dx}$(lnx) = $\frac{1}{x}$

(a)

Given

y = $log_{a}$$\sqrt{(1+x)}$

$\frac{dy}{dx}$ = $\frac{1}{lna\sqrt{(1+x)} }$ × $\frac{d}{dx}$ ($(1+x)^{\frac{1}{2} }$

   = $\frac{1}{lna\sqrt{(1+x)} }$ × $\frac{1}{2}$ $(1+x)^{-\frac{1}{2} }$ × $\frac{d}{dx}$ (1 + x)

   = $\frac{1}{lna\sqrt{(1+x)} }$ × $\frac{1}{2\sqrt{(1+x)} }$ × 1

   = $\frac{1}{2lna(1+x)}$

   = $\frac{1}{(1+x)lna^2}$

(b)

Given

y = ln sinx

$\frac{dy}{dx}$ = $\frac{1}{sinx}$ × $\frac{d}{dx}$(sinx)

   = $\frac{1}{sinx}$ × cosx

   = $\frac{cosx}{sinx}$

   = cotx