Answer:
see explanation
Step-by-step explanation:
Using the chain rule and the standard derivatives
Given
y = f(g(x)) , then
[tex]\frac{dy}{dx}[/tex] = f'(g(x)) × g'(x) ← chain rule
[tex]\frac{d}{dx}[/tex] (tanx) = sec²x , [tex]\frac{d}{dx}[/tex] (cotx) = - csc²x
(c)
y = tan[tex]\sqrt{x}[/tex] = tan[tex]x^{\frac{1}{2} }[/tex]
[tex]\frac{dy}{dx}[/tex] = sec²[tex]\sqrt{x}[/tex] × [tex]\frac{d}{dx}[/tex] ([tex]x^{\frac{1}{2} }[/tex] )
= sec²[tex]\sqrt{x}[/tex] × [tex]\frac{1}{2}[/tex][tex]x^{-\frac{1}{2} }[/tex]
= sec²[tex]\sqrt{x}[/tex] × [tex]\frac{1}{2x^{\frac{1}{2} } }[/tex]
= [tex]\frac{sec^2\sqrt{x} }{2\sqrt{x} }[/tex]
(d)
y = cot(1 + x)
[tex]\frac{dy}{dx}[/tex] = - csc²(1 + x) × [tex]\frac{d}{dx}[/tex] (1 + x)
= - csc²(1 + x) × 1
= - csc²(1 + x)
