Question

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

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the chain rule and the standard derivatives

Given

y = f(g(x)) , then

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

$\frac{d}{dx}$ (tanx) = sec²x , $\frac{d}{dx}$ (cotx) = - csc²x

(c)

y = tan$\sqrt{x}$ = tan$x^{\frac{1}{2} }$

$\frac{dy}{dx}$ = sec²$\sqrt{x}$ × $\frac{d}{dx}$ ($x^{\frac{1}{2} }$ )

    = sec²$\sqrt{x}$ × $\frac{1}{2}$$x^{-\frac{1}{2} }$

    = sec²$\sqrt{x}$ × $\frac{1}{2x^{\frac{1}{2} } }$

   = $\frac{sec^2\sqrt{x} }{2\sqrt{x} }$

(d)

y = cot(1 + x)

$\frac{dy}{dx}$ = - csc²(1 + x) × $\frac{d}{dx}$ (1 + x)

   = - csc²(1 + x) × 1

   = - csc²(1 + x)