1. Find the derivative of 1/cos(x).
Answer
1. sin(x)/cos2(x)
Solution
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).
2. Find the derivative of tan(x)
Answer
1. 1/cos2(x)
Solution
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).
3. Derive this formula for derivative of a quotient: (f/g)’= [f’g – fg’]/g2.
Answer
1. Hint: use the product rule on f/g = f ⋅ 1/g.
Solution
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.
4. Find the derivative of e(x)2.
Answer
1. 2x e(x)2
Solution
1. We will use the chain rule, with g = x2, f = eg: (eg)’ = eg ⋅ g’= eg ⋅ 2x = 2x e(x)2
5. Find the derivative of ln(cos(x2).
Answer
1. -2x tan (x2)
Solution
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)