Question

Hi. I need help with these questions (see image)
Please show workings.
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Answer 1

see explanation

Explanation:

Using the chain rule

Given

y = f(g(x)), then

`(dy)/(dx)` = f'(g(x)) × g'(x) ← chain rule

and the standard derivatives

`(d)/(dx)` (
`log_(a)` x ) =
`(1)/(xlna)` ,
`(d)/(dx)`(lnx) =
`(1)/(x)`

(a)

Given

y =
`log_(a)`
`√((1+x))`

`(dy)/(dx)` =
`(1)/(lna√((1+x)) )` ×
`(d)/(dx)` (
`(1+x)^{(1)/(2) }`

=
`(1)/(lna√((1+x)) )` ×
`(1)/(2)`
`(1+x)^{-(1)/(2) }` ×
`(d)/(dx)` (1 + x)

=
`(1)/(lna√((1+x)) )` ×
`(1)/(2√((1+x)) )` × 1

=
`(1)/(2lna(1+x))`

=
`(1)/((1+x)lna^2)`

(b)

Given

y = ln sinx

`(dy)/(dx)` =
`(1)/(sinx)` ×
`(d)/(dx)`(sinx)

=
`(1)/(sinx)` × cosx

=
`(cosx)/(sinx)`

= cotx