Example 1 – Higher Order Derivatives

Practice Questions with Step by Step Solutions

Question 1

Find [\dfrac{d^{4}}{dx^{4}}\big(\sin x\big)].

Step-by-step Solution:

1. [\dfrac{d}{dx}(\sin x)=\cos x].
2. [\dfrac{d^{2}}{dx^{2}}(\sin x)][=\dfrac{d}{dx}(\cos x) = -\sin x].
3. [\dfrac{d^{3}}{dx^{3}}(\sin x)][=\dfrac{d}{dx}(-\sin x) = -\cos x].
4. [\dfrac{d^{4}}{dx^{4}}(\sin x)][=\dfrac{d}{dx}(-\cos x) = \sin x].

Conclusion:
[\dfrac{d^{4}}{dx^{4}}\big(\sin x\big) ][= \sin x].

Question 2

Find [\dfrac{d^{3}}{dx^{3}}\big(\cos 2x\big)].

Step-by-step Solution:

1. [\dfrac{d}{dx}(\cos 2x) = -2\sin 2x] (chain rule).
2. [\dfrac{d^{2}}{dx^{2}}(\cos 2x) ][= \dfrac{d}{dx}(-2\sin 2x) ][= -2\cdot 2\cos 2x ][= -4\cos 2x].
3. [\dfrac{d^{3}}{dx^{3}}(\cos 2x) ][= \dfrac{d}{dx}(-4\cos 2x) ][= -4\cdot(-2\sin 2x) ][= 8\sin 2x].

Conclusion:
[\dfrac{d^{3}}{dx^{3}}\big(\cos 2x\big) = 8\sin 2x].

Question 3

Give a formula for [\dfrac{d^{n}}{dx^{n}}\big(\sin 3x\big)] and compute it for [n=5].

Step-by-step Solution:

1. General pattern: each differentiation brings a factor [3] (from chain rule) and rotates sine → cos → −sin → −cos → sine. So:
   [\dfrac{d^{n}}{dx^{n}}\big(\sin 3x\big) ][= 3^{n}\sin\big(3x + \tfrac{n\pi}{2}\big)].
2. For [n=5]: use formula or compute stepwise: factor [3^{5}=243] and shift by [5\pi/2] which is equivalent to [\pi/2] modulo [2\pi]. So:
   [\dfrac{d^{5}}{dx^{5}}\big(\sin 3x\big) ][= 3^{5}\sin\big(3x + \tfrac{5\pi}{2}\big) ][= 243\cos 3x].

(You can also follow the four-step cycle: n=5 same as n=1 with extra factor [3^{5}].)

Conclusion:
[\dfrac{d^{5}}{dx^{5}}\big(\sin 3x\big) ][= 243\cos 3x].

Question 4

Find [\dfrac{d^{2}}{dx^{2}}\big(\tan x\big)].

Step-by-step Solution:

1. [\dfrac{d}{dx}(\tan x) = \sec^{2}x].
2. [\dfrac{d^{2}}{dx^{2}}(\tan x) ][= \dfrac{d}{dx}(\sec^{2}x) ][= 2\sec x\cdot \dfrac{d}{dx}(\sec x)] (use chain/product view).
3. [\dfrac{d}{dx}(\sec x)][=\sec x\tan x], so substitute:
   [\dfrac{d^{2}}{dx^{2}}(\tan x) ][= 2\sec x\cdot (\sec x\tan x) ][= 2\sec^{2}x\tan x].

Conclusion:
[\dfrac{d^{2}}{dx^{2}}\big(\tan x\big) ][= 2\sec^{2}x\tan x].

Question 5

Find [\dfrac{d^{3}}{dx^{3}}\big(\sin^{2}x\big)].

Step-by-step Solution:

1. Recognize [\sin^{2}x = (\sin x)^{2}]. First derivative (chain rule):
   [\dfrac{d}{dx}(\sin^{2}x) ][= 2\sin x\cdot \cos x = \sin 2x].
2. Second derivative: [\dfrac{d^{2}}{dx^{2}}(\sin^{2}x) ][= \dfrac{d}{dx}(\sin 2x) ][= 2\cos 2x].
3. Third derivative: [\dfrac{d^{3}}{dx^{3}}(\sin^{2}x) ][= \dfrac{d}{dx}(2\cos 2x) ][= 2\cdot(-2\sin 2x) ][= -4\sin 2x].

Conclusion:
[\dfrac{d^{3}}{dx^{3}}\big(\sin^{2}x\big) ][= -4\sin 2x].

Question 6

Find [\dfrac{d}{dx}\big(\sec x \tan x\big)].

Step-by-step Solution:

1. Use product rule: if [u=\sec x], [v=\tan x] then [u’=\sec x\tan x], [v’=\sec^{2}x].
2. [\dfrac{d}{dx}(\sec x\tan x) = u’v + uv’ ][= (\sec x\tan x)\cdot \tan x + \sec x\cdot \sec^{2}x].
3. Simplify:
   [\dfrac{d}{dx}(\sec x\tan x) ][= \sec x\tan^{2}x + \sec^{3}x] ][= \sec x\big(\tan^{2}x + \sec^{2}x\big).

(You may leave it as the sum or factor [\sec x].)

Conclusion:
[\dfrac{d}{dx}\big(\sec x\tan x\big) ][= \sec x\tan^{2}x + \sec^{3}x].

Question 7

Find [\dfrac{d^{2}}{dx^{2}}\big(\dfrac{\sin x}{x}\big)], for [x\ne 0].

Step-by-step Solution:

1. First derivative using quotient rule: numerator [u=\sin x], denominator [v=x], [u’=\cos x], [v’=1].
   [\dfrac{d}{dx}\Big(\dfrac{\sin x}{x}\Big) ][= \dfrac{x\cos x – \sin x}{x^{2}}].
2. Second derivative: differentiate [\dfrac{x\cos x – \sin x}{x^{2}}]. Use quotient or write as [x^{-2}(x\cos x – \sin x)]. We’ll use product rule with [w = x^{-2}] and [z = x\cos x – \sin x].
3. Compute derivatives:
   [\dfrac{dw}{dx} = -2x^{-3}],
   [\dfrac{dz}{dx} ][= \dfrac{d}{dx}(x\cos x) – \cos x = (\cos x – x\sin x) – \cos x ][= -x\sin x].
   (Reason: [d(x cos x)][=cos x – x sin x], then subtract derivative of sin x which is cos x.)
4. Now [\dfrac{d}{dx}(wz) ][= w’ z + w z’ ][= -2x^{-3}(x\cos x – \sin x) + x^{-2}(-x\sin x)].
5. Simplify terms:
   First term = [-2\cdot \dfrac{x\cos x – \sin x}{x^{3}} ][= -\dfrac{2x\cos x – 2\sin x}{x^{3}}].
   Second term = [-\dfrac{x\sin x}{x^{2}} ][= -\dfrac{\sin x}{x}]. Put over common denominator [x^{3}]:
   [-\dfrac{2x\cos x – 2\sin x}{x^{3}} – \dfrac{\sin x}{x} ][= -\dfrac{2x\cos x – 2\sin x}{x^{3}} – \dfrac{\sin x , x^{2}}{x^{3}} ].
   Combine numerator: [-(2x\cos x – 2\sin x) – x^{2}\sin x] / [x^{3}] ][= [ -2x\cos x + 2\sin x – x^{2}\sin x ] / [x^{3}].
   We can write final simplified form as:
   [\dfrac{d^{2}}{dx^{2}}\Big(\dfrac{\sin x}{x}\Big) ][= -\dfrac{2x\cos x + x^{2}\sin x – 2\sin x}{x^{3}}].
   (You may reorder signs: multiply numerator by -1 accordingly.)

Conclusion:
[\dfrac{d^{2}}{dx^{2}}\Big(\dfrac{\sin x}{x}\Big) ][= -\dfrac{2x\cos x + x^{2}\sin x – 2\sin x}{x^{3}},][\quad] [x\ne 0].

Question 8

Find [\dfrac{d^{3}}{dx^{3}}\big(\cos x\big)].

Step-by-step Solution:

1. [\dfrac{d}{dx}(\cos x) = -\sin x].
2. [\dfrac{d^{2}}{dx^{2}}(\cos x) ][= \dfrac{d}{dx}(-\sin x) ][= -\cos x].
3. [\dfrac{d^{3}}{dx^{3}}(\cos x) ][= \dfrac{d}{dx}(-\cos x) ][= \sin x].

Conclusion:
[\dfrac{d^{3}}{dx^{3}}\big(\cos x\big) ][= \sin x].

Question 9

Find [\dfrac{d^{2}}{dx^{2}}\big(\cot x\big)].

Step-by-step Solution:

1. [\dfrac{d}{dx}(\cot x) = -\csc^{2}x].
2. Differentiate again: [\dfrac{d^{2}}{dx^{2}}(\cot x) ][= \dfrac{d}{dx}(-\csc^{2}x) ][= -2\csc x\cdot \dfrac{d}{dx}(\csc x)].
3. [\dfrac{d}{dx}(\csc x) ][= -\csc x\cot x], so substitute:
   [\dfrac{d^{2}}{dx^{2}}(\cot x) ][= -2\csc x\cdot(-\csc x\cot x) ][= 2\csc^{2}x\cot x].

Conclusion:
[\dfrac{d^{2}}{dx^{2}}\big(\cot x\big) ][= 2\csc^{2}x\cot x].

Question 10

Find [\dfrac{d^{4}}{dx^{4}}\big(\sin 3x + \cos 2x\big)].

Step-by-step Solution:

1. Differentiate each term separately and use linearity. For [\sin 3x], each derivative multiplies by [3] and cycles; for [\cos 2x], multiply by [2] and cycle. Fourth derivative brings factor [3^{4}] for [\sin 3x] and [2^{4}] for [\cos 2x], with original function sign preserved because 4 is multiple of 4. Formally:
   [\dfrac{d^{4}}{dx^{4}}(\sin 3x) ][= 3^{4}\sin 3x ][= 81\sin 3x].
   [\dfrac{d^{4}}{dx^{4}}(\cos 2x) ][= 2^{4}\cos 2x ][= 16\cos 2x].
2. Add results:

Conclusion:
[\dfrac{d^{4}}{dx^{4}}\big(\sin 3x + \cos 2x\big) ][= 81\sin 3x + 16\cos 2x].