Answer:
see explanation
Step-by-step explanation:
Using the chain rule and the standard derivatives
Given
y = f(g(x)) , then
= f'(g(x)) × g'(x) ← chain rule
(tanx) = sec²x ,
(cotx) = - csc²x
(c)
y = tan
= tan
= sec²
×
(
)
= sec²
× ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
![x^{-\frac{1}{2} }](https://tex.z-dn.net/?f=x%5E%7B-%5Cfrac%7B1%7D%7B2%7D%20%7D)
= sec²
× ![\frac{1}{2x^{\frac{1}{2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2x%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%7D)
= ![\frac{sec^2\sqrt{x} }{2\sqrt{x} }](https://tex.z-dn.net/?f=%5Cfrac%7Bsec%5E2%5Csqrt%7Bx%7D%20%7D%7B2%5Csqrt%7Bx%7D%20%7D)
(d)
y = cot(1 + x)
= - csc²(1 + x) ×
(1 + x)
= - csc²(1 + x) × 1
= - csc²(1 + x)
The answer is pq^2, for it can be taken out of both monomials.
Answer: Spain, France, Britain
Step-by-step explanation:
2y^2 (3x+5z)
6xy^2 + 10zy^2