Example 1 – Differentiation of Function in the form f(x) by g(x)

Question 1.

Differentiate [f(x)][=\dfrac{x^{2}+1}{x+1}] , [x\ne -1].

Step-by-Step Solution:

1. Let [u=x^{2}+1], [v=x+1].
2. Compute derivatives: [u’=2x], [v’=1].
3. Apply quotient rule:
   [\dfrac{v u’ – u v’}{v^{2}}][ = \dfrac{(x+1)(2x) – (x^{2}+1)(1)}{(x+1)^{2}}].
4. Expand numerator: [(2x^{2}+2x) – (x^{2}+1)][ = x^{2}+2x-1].
5. Final simplified derivative:
   [\dfrac{x^{2}+2x-1}{(x+1)^{2}}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{x^{2}+1}{x+1}\big)][=\dfrac{x^{2}+2x-1}{(x+1)^{2}},][\quad][ x\ne -1.]

Question 2.

Differentiate [f(x)=\dfrac{3x-2}{x^{2}+1}] .

Step-by-Step Solution:

1. Let [u=3x-2], [v=x^{2}+1].
2. Derivatives: [u’=3], [v’=2x].
3. Quotient rule:
   [\dfrac{(x^{2}+1)(3) – (3x-2)(2x)}{(x^{2}+1)^{2}}].
4. Expand numerator: [3x^{2}+3 – (6x^{2}-4x)][ = 3x^{2}+3 -6x^{2}+4x][ = -3x^{2}+4x+3].
5. Final form:
   [\dfrac{-3x^{2}+4x+3}{(x^{2}+1)^{2}}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{3x-2}{x^{2}+1}\big)][=\dfrac{-3x^{2}+4x+3}{(x^{2}+1)^{2}}.]

Question 3.

Differentiate [f(x)=\dfrac{\sin x}{x}] , [x\ne 0].

Step-by-Step Solution:

1. Let [u=\sin x], [v=x].
2. Derivatives: [u’=\cos x], [v’=1].
3. Quotient rule:
   [\dfrac{x\cos x – \sin x\cdot 1}{x^{2}}].
4. Final simplified derivative:
   [\dfrac{x\cos x – \sin x}{x^{2}}].

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

Question 4.

Differentiate [f(x)=\dfrac{e^{x}}{x}] , [x\ne 0].

Step-by-Step Solution:

1. Let [u=e^{x}], [v=x].
2. Derivatives: [u’=e^{x}], [v’=1].
3. Quotient rule:
   [\dfrac{x e^{x} – e^{x}\cdot 1}{x^{2}}][ = \dfrac{e^{x}(x-1)}{x^{2}}].
4. Final derivative written neatly:
   [\dfrac{e^{x}(x-1)}{x^{2}}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{e^{x}}{x}\big)][=\dfrac{e^{x}(x-1)}{x^{2}}][,\quad][ x\ne 0.]

Question 5.

Differentiate [f(x)=\dfrac{\ln x}{x^{2}}] , [x>0].

Step-by-Step Solution:

1. Let [u=\ln x], [v=x^{2}].
2. Derivatives: [u’=\dfrac{1}{x}], [v’=2x].
3. Quotient rule:
   [\dfrac{x^{2}\cdot\dfrac{1}{x} – \ln x \cdot 2x}{x^{4}}].
4. Divide by [x^{4}]: final derivative = [\dfrac{x(1-2\ln x)}{x^{4}}][ = \dfrac{1-2\ln x}{x^{3}}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{\ln x}{x^{2}}\big)][=\dfrac{1-2\ln x}{x^{3}},][\quad][ x>0.]

Question 6.

Differentiate [f(x)=\dfrac{x}{1+x^{2}}].

Step-by-Step Solution:

1. Let [u=x], [v=1+x^{2}].
2. Derivatives: [u’=1], [v’=2x].
3. Quotient rule:
   [\dfrac{(1+x^{2})\cdot 1 – x\cdot 2x}{(1+x^{2})^{2}}].
4. Simplify numerator: [1+x^{2} – 2x^{2}][ = 1 – x^{2}].
5. Final derivative:
   [\dfrac{1-x^{2}}{(1+x^{2})^{2}}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{x}{1+x^{2}}\big)][=\dfrac{1-x^{2}}{(1+x^{2})^{2}}.]

Question 7.

Differentiate [f(x)=\dfrac{\tan x}{x}] , ([x\ne 0] and where tan defined).

Step-by-Step Solution:

1. Let [u=\tan x], [v=x].
2. Derivatives: [u’=\sec^{2}x], [v’=1].
3. Quotient rule:
   [\dfrac{x\sec^{2}x – \tan x\cdot 1}{x^{2}}].
4. Final derivative:
   [\dfrac{x\sec^{2}x – \tan x}{x^{2}}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{\tan x}{x}\big)][=\dfrac{x\sec^{2}x – \tan x}{x^{2}},][\quad][ x\ne 0.]

Question 8.

Differentiate [f(x)=\dfrac{x^{3}+2}{x^{2}}] , [x\ne 0].

Step-by-Step Solution:

1. Option A: Use quotient rule directly. (We will show quotient rule method.)
   Let [u=x^{3}+2], [v=x^{2}].
   Derivatives: [u’=3x^{2}], [v’=2x].
   Quotient rule:
   [\dfrac{x^{2}\cdot 3x^{2} – (x^{3}+2)\cdot 2x}{x^{4}}].
   Simplify numerator: [3x^{4} – 2x^{4} – 4x][ = x^{4} – 4x].
   Divide: [\dfrac{x^{4}-4x}{x^{4}}][ = 1 – \dfrac{4}{x^{3}}].
2. Option B (shortcut): simplify first: [\dfrac{x^{3}+2}{x^{2}}][ = x + 2x^{-2}]. Then derivative = [1 – 4x^{-3}] [= 1 – \dfrac{4}{x^{3}}]. (Same result.)

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{x^{3}+2}{x^{2}}\big)][=1 – \dfrac{4}{x^{3}},][\quad][ x\ne 0.]

Question 9.

Differentiate [f(x)=\dfrac{\sec x}{x}] , [x\ne 0] and where sec defined.

Step-by-Step Solution:

1. Let [u=\sec x], [v=x].
2. Derivatives: [u’=\sec x\tan x], [v’=1].
3. Quotient rule:
   [\dfrac{x\cdot (\sec x\tan x) – \sec x\cdot 1}{x^{2}}].
4. Factor [\sec x] in numerator: [\dfrac{\sec x (x\tan x – 1)}{x^{2}}].
5. Final derivative:
   [\dfrac{\sec x (x\tan x – 1)}{x^{2}}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{\sec x}{x}\big)][=\dfrac{\sec x (x\tan x – 1)}{x^{2}}][,\quad][ x\ne 0.]

Question 10.

Differentiate [f(x)=\dfrac{\tan^{-1}x}{x}] , [x\ne 0].

Step-by-Step Solution:

1. Let [u=\tan^{-1}x], [v=x].
2. Derivatives: [u’=\dfrac{1}{1+x^{2}}], [v’=1].
3. Quotient rule:
   [\dfrac{x\cdot \dfrac{1}{1+x^{2}} – \tan^{-1}x \cdot 1}{x^{2}}].
4. Combine numerator over common denominator [1+x^{2}] if desired:
   Numerator = [\dfrac{x}{1+x^{2}} – \tan^{-1}x]. To write over common denom: [\dfrac{x – (1+x^{2})\tan^{-1}x}{1+x^{2}}]. Then whole expression:
   [\dfrac{\dfrac{x – (1+x^{2})\tan^{-1}x}{1+x^{2}}}{x^{2}}][ = \dfrac{x – (1+x^{2})\tan^{-1}x}{x^{2}(1+x^{2})}].
5. Either form acceptable; a clear simplified final form:
   [\dfrac{x}{x^{2}(1+x^{2})} – \dfrac{\tan^{-1}x}{x^{2}}] or compact: [\dfrac{x – (1+x^{2})\tan^{-1}x}{x^{2}(1+x^{2})}].

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{\tan^{-1}x}{x}\big)][=\dfrac{x – (1+x^{2})\tan^{-1}x}{x^{2}(1+x^{2})},][\quad][ x\ne 0.]