Differentiation by Applying all Rules

Practice Questions (with Step-by-Step Solutions)

Question 1.

Differentiate [f(x)=x^{3} + 4x^{2} – 5x + 7].

Step-by-Step Solution:

1. This is a sum of terms: [x^{3}], [4x^{2}], [ -5x], [7].
2. Differentiate each term using power & constant rules:
   • [\dfrac{d}{dx}(x^{3}) = 3x^{2}]
   • [\dfrac{d}{dx}(4x^{2})][ = 4\cdot 2x = 8x]
   • [\dfrac{d}{dx}(-5x) = -5]
   • [\dfrac{d}{dx}(7) = 0]
3. Add the results (linearity): [3x^{2} + 8x – 5].

Conclusion:
[\dfrac{d}{dx}\big(x^{3}+4x^{2}-5x+7\big)][ = 3x^{2} + 8x – 5].

Question 2.

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

Step-by-Step Solution:

1. Recognize sum of two parts: [x\sin x] (product) and [\cos x].
2. Differentiate [x\sin x] using product rule: let [u=x], [v=\sin x]; then [u’=1], [v’=\cos x].
   • Product rule gives: [u v’ + v u’ = x\cos x + \sin x].
3. Differentiate [\cos x]: [\dfrac{d}{dx}(\cos x) = -\sin x].
4. Combine results: [x\cos x + \sin x – \sin x]. The [\sin x] terms cancel.

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

Question 3.

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

Step-by-Step Solution:

1. Identify numerator [u=x^{2}+1] and denominator [v=x-1].
2. Compute derivatives: [u’ = 2x], [v’ = 1].
3. Apply quotient rule:
   [\dfrac{d}{dx}\big(\dfrac{u}{v}\big)][ = \dfrac{v u’ – u v’}{v^{2}}]
   Substitute: [\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 form: [\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 4.

Differentiate [f(x)=(3x^{2}+2x) e^{x}].

Step-by-Step Solution:

1. This is a product: [u=3x^{2}+2x], [v=e^{x}].
2. Compute derivatives: [u’ = 6x+2], [v’ = e^{x}].
3. Apply product rule: [u v’ + v u’ = (3x^{2}+2x)e^{x} + e^{x}(6x+2)].
4. Factor common [e^{x}]: [e^{x}\big(3x^{2}+2x + 6x+2\big)][ = e^{x}\big(3x^{2}+8x+2\big)].

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

Question 5.

Differentiate [f(x)=\dfrac{\sin x}{1+x^{2}}], [x\in\mathbb{R}].

Step-by-Step Solution:

1. Numerator [u=\sin x], denominator [v=1+x^{2}].
2. Derivatives: [u’=\cos x], [v’=2x].
3. Quotient rule: [\dfrac{v u’ – u v’}{v^{2}}][ = \dfrac{(1+x^{2})\cos x – \sin x(2x)}{(1+x^{2})^{2}}].
4. Optionally leave as is; this is a clean final form.

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

Question 6.

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

Step-by-Step Solution:

1. This is a composition: outer [F(u)=u^{3}], inner [g(x)=2x+1].
2. [\dfrac{d}{du}(u^{3}) = 3u^{2}].
3. [\dfrac{d}{dx}(2x+1) = 2].
4. Chain rule: [3(2x+1)^{2}\cdot 2 = 6(2x+1)^{2}].

Conclusion:
[\dfrac{d}{dx}\big((2x+1)^{3}\big)][ = 6(2x+1)^{2}].

Question 7.

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

Step-by-Step Solution:

1. Product of [u=x^{2}] and [v=\ln x].
2. Derivatives: [u’ = 2x], [v’ = \dfrac{1}{x}].
3. Product rule: [u v’ + v u’ = x^{2}\cdot \dfrac{1}{x} + \ln x \cdot 2x].
4. Simplify first term: [x^{2}\cdot \dfrac{1}{x} = x]. So derivative = [x + 2x\ln x]. Factor if desired: [x(1+2\ln x)].

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

Question 8.

Differentiate [f(x)=\dfrac{e^{x}}{\sqrt{1+x^{2}}}], domain [1+x^{2}>0].

Step-by-Step Solution:

1. Write denominator as [v=(1+x^{2})^{1/2}]. Numerator [u=e^{x}].
2. Derivatives: [u’ = e^{x}], [v’ = \dfrac{1}{2}(1+x^{2})^{-1/2}\cdot 2x][ = \dfrac{x}{\sqrt{1+x^{2}}}].
3. Quotient rule: [\dfrac{v u’ – u v’}{v^{2}}]
   Substitute: [\dfrac{(1+x^{2})^{1/2}\cdot e^{x} – e^{x}\cdot \dfrac{x}{\sqrt{1+x^{2}}}}{(1+x^{2})}].
4. Factor [e^{x}] and common denominator: rewrite numerator over [\sqrt{1+x^{2}}]:
   Numerator = [e^{x}\big( (1+x^{2}) – x \big)/\sqrt{1+x^{2}} ]
   So whole fraction = [\dfrac{e^{x}\big(1+x^{2}-x\big)}{(1+x^{2})\sqrt{1+x^{2}}}]
   Alternatively simplify step-by-step: keep a tidy final form:
   [\dfrac{e^{x}\big( (1+x^{2}) – x \big)}{(1+x^{2})^{3/2}}].
5. Final simplified derivative: [\dfrac{e^{x}(1 + x^{2} – x)}{(1+x^{2})^{3/2}}].

Conclusion:
[\dfrac{d}{dx}\Big(\dfrac{e^{x}}{\sqrt{1+x^{2}}}\Big)][ = \dfrac{e^{x}\big(1 + x^{2} – x\big)}{(1+x^{2})^{3/2}}].

Question 9.

Differentiate [f(x)=\sin\big(x^{2}\big)\cdot x].

Step-by-Step Solution:

1. This is a product: [u = \sin(x^{2})], [v = x]. Note [u] itself is a composition requiring chain rule.
2. Compute [u’] using chain rule: inner [g(x)=x^{2}], outer [F(u)=\sin u].
   • [\dfrac{d}{du}(\sin u)=\cos u], [\dfrac{d}{dx}(x^{2})=2x].
   • So [u’ = \cos(x^{2})\cdot 2x = 2x\cos(x^{2})].
3. Compute [v’]: [1].
4. Apply product rule: [u’v + uv’ = (2x\cos(x^{2}))\cdot x + \sin(x^{2})\cdot 1].
5. Simplify first term: [2x\cos(x^{2})\cdot x = 2x^{2}\cos(x^{2})]. Combine: [2x^{2}\cos(x^{2}) + \sin(x^{2})].

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

Question 10.

Differentiate [f(x)=\dfrac{(x+1)^{2}}{ \cos x }], domain where [\cos x \ne 0].

Step-by-Step Solution:

1. Recognize quotient with numerator [u=(x+1)^{2}] (composition) and denominator [v=\cos x].
2. Compute [u’] via chain rule: outer [U(t)=t^{2}], inner [t=x+1]; [U'(t)=2t], [t’ = 1] ⇒ [u’ = 2(x+1)].
3. Compute [v’]: [\dfrac{d}{dx}(\cos x) = -\sin x].
4. Apply quotient rule: [\dfrac{v u’ – u v’}{v^{2}}]
   Substitute: [\dfrac{\cos x \cdot 2(x+1) – (x+1)^{2}\cdot(-\sin x)}{\cos^{2} x}].
5. Simplify signs: numerator = [2(x+1)\cos x + (x+1)^{2}\sin x]. Factor common [x+1] if desired:
   Numerator = [(x+1)\big(2\cos x + (x+1)\sin x\big)]. So final form:
   [\dfrac{(x+1)\big(2\cos x + (x+1)\sin x\big)}{\cos^{2}x}].

Conclusion:
[\dfrac{d}{dx}\Big(\dfrac{(x+1)^{2}}{\cos x}\Big)][ = \dfrac{2(x+1)\cos x + (x+1)^{2}\sin x}{\cos^{2}x},] [\quad] [\cos x \ne 0]
(or the factored variant shown above).