Example 1 – Derivative of Sum and Difference of Two Functions

Question 1.

Differentiate [f(x)=x^{4} + 2x^{2} – 7x].

Step-by-Step Solution:

1. Identify terms: [u(x)=x^{4}], [v(x)=2x^{2}], [w(x)=-7x].
2. Differentiate each term using power rule:
   • [\dfrac{d}{dx}(x^{4})=4x^{3}]
   • [\dfrac{d}{dx}(2x^{2})=2\cdot 2x=4x]
   • [\dfrac{d}{dx}(-7x)=-7]
3. Use linearity (sum/difference): add derivatives: [4x^{3}+4x-7].

Conclusion:
[\dfrac{d}{dx}(x^{4}+2x^{2}-7x)][=4x^{3}+4x-7].

Question 2.

Differentiate [f(x)=e^{2x} + \ln x] (domain: [x>0]).

Step-by-Step Solution:

1. Split: [u(x)=e^{2x}], [v(x)=\ln x].
2. Differentiate [u] using chain rule: derivative of [e^{2x}] is [e^{2x}\cdot \dfrac{d}{dx}(2x)=2e^{2x}].
3. Differentiate [v]: [\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}].
4. Sum the results: [2e^{2x}+\dfrac{1}{x}].

Conclusion:
[\dfrac{d}{dx}\big(e^{2x}+\ln x\big)][=2e^{2x}+\dfrac{1}{x}], valid for [x>0].

Question 3.

Differentiate [f(x)=\sin x – \cos x + x].

Step-by-Step Solution:

1. Identify parts: [u=\sin x], [v=-\cos x], [w=x].
2. Differentiate each:
   • [\dfrac{d}{dx}(\sin x)=\cos x]
   • [\dfrac{d}{dx}(-\cos x)=-(-\sin x)][=+\sin x] (careful: derivative of [\cos x] is [ -\sin x ], so minus that gives [+\sin x])
   • [\dfrac{d}{dx}(x)=1]
3. Add them: [\cos x + \sin x + 1].

Conclusion:
[\dfrac{d}{dx}(\sin x – \cos x + x)][=\cos x + \sin x + 1].

Question 4.

Differentiate [f(x)=\tan x + \sec x].

Step-by-Step Solution:

1. Terms: [u=\tan x], [v=\sec x].
2. Use standard derivatives:
   • [\dfrac{d}{dx}(\tan x)=\sec^{2}x]
   • [\dfrac{d}{dx}(\sec x)=\sec x \tan x]
3. Sum: [\sec^{2}x + \sec x \tan x].

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

Question 5.

Differentiate [f(x)=\sqrt{x} + x^{-3}] (domain: [x>0]).

Step-by-Step Solution:

1. Rewrite: [\sqrt{x}=x^{1/2}], other term [x^{-3}].
2. Differentiate using power rule:
   • [\dfrac{d}{dx}(x^{1/2})][=\dfrac{1}{2}x^{-1/2}][=\dfrac{1}{2\sqrt{x}}]
   • [\dfrac{d}{dx}(x^{-3})=-3x^{-4}][=-\dfrac{3}{x^{4}}]
3. Add: [\dfrac{1}{2}x^{-1/2} – 3x^{-4}].

Conclusion:
[\dfrac{d}{dx}\Big(\sqrt{x}+x^{-3}\Big)][=\dfrac{1}{2\sqrt{x}} – \dfrac{3}{x^{4}}], for [x>0].

Question 6.

Differentiate [f(x)][=\dfrac{1}{x} + \ln(x^{2}+1)] (domain: [x\ne 0], but ln term ok for all real).

Step-by-Step Solution:

1. Terms: [u=\dfrac{1}{x}=x^{-1}], [v=\ln(x^{2}+1)].
2. Differentiate [u]: [\dfrac{d}{dx}(x^{-1})][=-x^{-2}][=-\dfrac{1}{x^{2}}].
3. Differentiate [v] using chain rule: inner [g(x)=x^{2}+1], [g'(x)=2x]; so [\dfrac{d}{dx}\ln(g(x))][=\dfrac{g'(x)}{g(x)}][=\dfrac{2x}{x^{2}+1}].
4. Sum: [ -\dfrac{1}{x^{2}}][ + \dfrac{2x}{x^{2}+1} ].

Conclusion:
[\dfrac{d}{dx}\Big(\dfrac{1}{x}+\ln(x^{2}+1)\Big)][ = -\dfrac{1}{x^{2}} + \dfrac{2x}{x^{2}+1}], valid where defined.

Question 7.

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

Step-by-Step Solution:

1. Terms: [u=\tan^{-1}x], [v=x^{3}].
2. Standard derivatives:
   • [\dfrac{d}{dx}(\tan^{-1}x)][=\dfrac{1}{1+x^{2}}]
   • [\dfrac{d}{dx}(x^{3})=3x^{2}]
3. Add: [\dfrac{1}{1+x^{2}} + 3x^{2}].

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

Question 8.

Differentiate [f(x)=a^{x} + \sin x], where [a>0] is a constant.

Step-by-Step Solution:

1. Terms: [u=a^{x}], [v=\sin x].
2. Standard derivatives:
   • [\dfrac{d}{dx}(a^{x})=a^{x}\ln a]
   • [\dfrac{d}{dx}(\sin x)=\cos x]
3. Sum: [a^{x}\ln a + \cos x].

Conclusion:
[\dfrac{d}{dx}(a^{x}+\sin x)][=a^{x}\ln a + \cos x].

Question 9.

Differentiate [f(x)][=\sec^{-1}x + \csc^{-1}x] (domain: [|x|>1]).

Step-by-Step Solution:

1. Terms: [u=\sec^{-1}x], [v=\csc^{-1}x].
2. Use standard derivatives (careful with signs):
   • [\dfrac{d}{dx}(\sec^{-1}x)][=\dfrac{1}{|x|\sqrt{x^{2}-1}}] (for [|x|>1])
   • [\dfrac{d}{dx}(\csc^{-1}x)][=-\dfrac{1}{|x|\sqrt{x^{2}-1}}] (for [|x|>1])
3. Add: they cancel: [\dfrac{1}{|x|\sqrt{x^{2}-1}}][ – \dfrac{1}{|x|\sqrt{x^{2}-1}} = 0].

Conclusion:
[\dfrac{d}{dx}\big(\sec^{-1}x+\csc^{-1}x\big)=0], for [|x|>1]. (Nice cancellation — derivative zero on domain.)

Question 10.

Differentiate [f(x)][=\sin^{-1}x + \sqrt{1-x^{2}}] (domain: [ -1<x<1 ]).

Step-by-Step Solution:

1. Terms: [u=\sin^{-1}x], [v=\sqrt{1-x^{2}}=(1-x^{2})^{1/2}].
2. Differentiate [u]: [\dfrac{d}{dx}(\sin^{-1}x)][=\dfrac{1}{\sqrt{1-x^{2}}}].
3. Differentiate [v] using chain rule:
   • Write [v=(1-x^{2})^{1/2}].
   • [\dfrac{d}{dx}(1-x^{2})^{1/2}][=\dfrac{1}{2}(1-x^{2})^{-1/2}\cdot \dfrac{d}{dx}(1-x^{2})]
   • [\dfrac{d}{dx}(1-x^{2}) = -2x]
   • So [\dfrac{d}{dx}(\sqrt{1-x^{2}})][=\dfrac{1}{2}(1-x^{2})^{-1/2}\cdot(-2x)][= -\dfrac{x}{\sqrt{1-x^{2}}}].
4. Add derivatives:
   [\dfrac{1}{\sqrt{1-x^{2}}} + \Big(-\dfrac{x}{\sqrt{1-x^{2}}}\Big)][ = \dfrac{1-x}{\sqrt{1-x^{2}}}.]

Conclusion:
[\dfrac{d}{dx}\Big(\sin^{-1}x + \sqrt{1-x^{2}}\Big)][=\dfrac{1-x}{\sqrt{1-x^{2}}}], valid for [ -1<x<1 ].