Question

Hi. I need help with these questions (see image)

Please show workings.
Answer c and d.
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Answer 1

see explanation

Explanation:

Using the chain rule and the standard derivatives

Given

y = f(g(x)) , then

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

`(d)/(dx)` (tanx) = sec²x ,
`(d)/(dx)` (cotx) = - csc²x

(c)

y = tan
`√(x)` = tan
`x^{(1)/(2) }`

`(dy)/(dx)` = sec²
`√(x)` ×
`(d)/(dx)` (
`x^{(1)/(2) }` )

= sec²
`√(x)` ×
`(1)/(2)`
`x^{-(1)/(2) }`

= sec²
`√(x)` ×
`\frac{1}{2x^{(1)/(2) } }`

=
`(sec^2√(x) )/(2√(x) )`

(d)

y = cot(1 + x)

`(dy)/(dx)` = - csc²(1 + x) ×
`(d)/(dx)` (1 + x)

= - csc²(1 + x) × 1

= - csc²(1 + x)