1. sin(x)/cos²(x)

1. We will use the chain rule, with g = cos(x), so f = 1/g. (1/g)’ = -1/g² ⋅ (-sin(x)) =sin(x)/cos²(x).

1. 1/cos²(x)

    1. We can write tan(x) = sin(x)/cos(x) = sin(x) ⋅ 1/cos(x). From the previous problem, we know the derivative of 1/cos(x), so we can use the product rule on this: (sin(x) ⋅ 1/cos(x))’ = cos(x) ⋅ 1/cos(x) + sin(x) ⋅ sin(x)/cos2(x) = 1 + sin²(x)/cos²(x) = [cos²(x) + sin²(x)] /cos²(x) = 1/cos²(x).

1. Hint: use the product rule on f/g = f ⋅ 1/g.

1. The approach here follows exactly the computation of tan(x)’ in the previous exercise. We will use the product rule, writing f/g = f ⋅ 1/g: (f ⋅ 1/g)’= f’ ⋅ 1/g + f ⋅ (1/g)’ Now we use the chain rule to find the derivative of 1/g, just like with 1/cos(x) in the first problem: (1/g)’= (-1/g²) ⋅ g’= – g’/g² Putting this into the calculation of (f/g)’, we get  (f/g)’= f’ ⋅ 1/g + f ⋅ (-g’/g²) = f’ / g – fg’ / g2 = [f’g – fg’]/g².

1. 2x e(^x)²

1. We will use the chain rule, with g = x², f = e^g: (e^g)’ = e^g ⋅ g’= e^g ⋅ 2x = 2x e(^x)²

1. -2x tan (x²)

1. We will start by using the chain rule with g = cos(x²), and f = ln(g).(ln(g))’ = 1/g ⋅ g’ To get the derivative of g, we will use chain rule again, with h = x2, and g = cos(h): (cos(h))’= -sin(h) ⋅ h’ = -sin(x²) ⋅ 2x = -2x sin(x²). Putting that back into our calculation, we get     (ln(g))’ = 1/g ⋅ [-2x sin(x²)] = -sin(x²)/cos(x²) = -2x tan(x²)