Example Set 1 – Logarithmic Differentiation

Practice Questions (with Step-by-Step Solutions)

Question 1.
Differentiate [y = x^{x}], [x>0].

Step-by-Step Solution:

1. Take natural log of both sides: [\ln y = \ln(x^{x})].
2. Use log-power rule: [\ln y = x\ln x].
3. Differentiate implicitly w.r.t [x]: [\dfrac{1}{y}\dfrac{dy}{dx}][ = \ln x + x\cdot\dfrac{1}{x}][ = \ln x + 1].
4. Multiply both sides by [y]: [\dfrac{dy}{dx} = y(\ln x + 1)].
5. Substitute back [y = x^{x}]: [\dfrac{dy}{dx} = x^{x}(\ln x + 1)].

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

Question 2.
Differentiate [y = (x^{2}+1)^{x}], [x>0].

Step-by-Step Solution:

1. Take ln: [\ln y = x\ln(x^{2}+1)].
2. Differentiate implicitly:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = 1\cdot \ln(x^{2}+1) + x\cdot \dfrac{1}{x^{2}+1}\cdot 2x]
   [= \ln(x^{2}+1) + \dfrac{2x^{2}}{x^{2}+1}].
3. Multiply by [y]:
   [\dfrac{dy}{dx} = (x^{2}+1)^{x}\Big(\ln(x^{2}+1) + \dfrac{2x^{2}}{x^{2}+1}\Big)].

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

Question 3.
Differentiate [y = x^{\sin x}], domain [x>0].

Step-by-Step Solution:

1. Take ln: [\ln y = \sin x \cdot \ln x].
2. Differentiate implicitly:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = \cos x\cdot \ln x + \sin x\cdot \dfrac{1}{x}].
3. Multiply by [y]:
   [\dfrac{dy}{dx}][ = x^{\sin x}\Big(\cos x\ln x + \dfrac{\sin x}{x}\Big)].

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

Question 4.
Differentiate [y = \dfrac{x^{2}\sqrt{x}}{(x+1)^{3}}], domain [x>0].

Step-by-Step Solution:

1. Write: [y][ = \dfrac{x^{2}\cdot x^{1/2}}{(x+1)^{3}}][ = \dfrac{x^{5/2}}{(x+1)^{3}}].
2. Take ln: [\ln y = \ln(x^{5/2}) – \ln((x+1)^{3})][ = \dfrac{5}{2}\ln x – 3\ln(x+1)].
3. Differentiate implicitly:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = \dfrac{5}{2}\cdot\dfrac{1}{x} – 3\cdot\dfrac{1}{x+1}].
4. Multiply by [y]:
   [\dfrac{dy}{dx}][ = \dfrac{x^{5/2}}{(x+1)^{3}}\Big(\dfrac{5}{2x} – \dfrac{3}{x+1}\Big)].
5. (Optional) Simplify inside the bracket to a single fraction if needed.

Conclusion:
[\dfrac{d}{dx}\Big(\dfrac{x^{2}\sqrt{x}}{(x+1)^{3}}\Big)][ = \dfrac{x^{5/2}}{(x+1)^{3}}\Big(\dfrac{5}{2x} – \dfrac{3}{x+1}\Big)], [x>0].

Question 5.
Differentiate [y = ( \sin x )^{\cos x}], domain where [\sin x>0].

Step-by-Step Solution:

1. Take ln: [\ln y = \cos x \cdot \ln(\sin x)].
2. Differentiate implicitly using product & chain rules:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = -\sin x \cdot \ln(\sin x) + \cos x \cdot \dfrac{\cos x}{\sin x}]
   (since [\dfrac{d}{dx}(\cos x)][ = -\sin x] and [\dfrac{d}{dx}\ln(\sin x)][=\dfrac{\cos x}{\sin x}]).
3. Multiply by [y]:
   [\dfrac{dy}{dx}][ = (\sin x)^{\cos x}\Big(-\sin x\ln(\sin x) + \dfrac{\cos^{2}x}{\sin x}\Big)].
4. (Optional) Combine terms over common denominator [\sin x] if desired.

Conclusion:
[\dfrac{d}{dx}\big((\sin x)^{\cos x}\big)][ = (\sin x)^{\cos x}\Big(-\sin x\ln(\sin x) + \dfrac{\cos^{2}x}{\sin x}\Big)], valid where [\sin x>0].

Question 6.
Differentiate [y = (x+1)^{x+2}], domain [x>-1].

Step-by-Step Solution:

1. Take ln: [\ln y = (x+2)\ln(x+1)].
2. Differentiate implicitly:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = 1\cdot\ln(x+1) + (x+2)\cdot\dfrac{1}{x+1}\cdot 1].
   [ = \ln(x+1) + \dfrac{x+2}{x+1} ].
3. Multiply by [y]:
   [\dfrac{dy}{dx}][ = (x+1)^{x+2}\Big(\ln(x+1) + \dfrac{x+2}{x+1}\Big)].

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

Question 7.
Differentiate [y][= \dfrac{(x^{2}+1)^{3}}{x^{x}}], domain [x>0].

Step-by-Step Solution:

1. Take ln:
   [\ln y][ = 3\ln(x^{2}+1) – \ln(x^{x}) = 3\ln(x^{2}+1) – x\ln x].
2. Differentiate implicitly:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = 3\cdot \dfrac{2x}{x^{2}+1} – \big(\ln x + x\cdot \dfrac{1}{x}\big)]
   [= \dfrac{6x}{x^{2}+1} – (\ln x + 1)].
3. Multiply by [y]:
   [\dfrac{dy}{dx}][ = \dfrac{(x^{2}+1)^{3}}{x^{x}}\Big(\dfrac{6x}{x^{2}+1} – \ln x – 1\Big)].
4. (Optional) Simplify factor [ (x^{2}+1) term ] inside if needed.

Conclusion:
[\dfrac{d}{dx}\Big(\dfrac{(x^{2}+1)^{3}}{x^{x}}\Big)][= \dfrac{(x^{2}+1)^{3}}{x^{x}}\Big(\dfrac{6x}{x^{2}+1} – \ln x – 1\Big)], [x>0].

Question 8.
Differentiate [y = \big( \tan^{-1}x \big)^{2}], domain all real.

Step-by-Step Solution:

1. Take ln: [\ln y = 2\ln(\tan^{-1}x)].
2. Differentiate implicitly:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = 2\cdot \dfrac{1}{\tan^{-1}x}\cdot \dfrac{1}{1+x^{2}}].
   (Since [\dfrac{d}{dx}\tan^{-1}x][ = \dfrac{1}{1+x^{2}}].)
3. Multiply by [y]:
   [\dfrac{dy}{dx}][ = (\tan^{-1}x)^{2}\cdot \dfrac{2}{\tan^{-1}x}\cdot \dfrac{1}{1+x^{2}}].
4. Simplify: one factor cancels:
   [\dfrac{dy}{dx}][ = \dfrac{2\tan^{-1}x}{1+x^{2}}].

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

Question 9.
Differentiate [y = (x^{x^{2}})], domain [x>0].

Step-by-Step Solution:

1. Take ln: [\ln y = x^{2}\ln x].
2. Differentiate implicitly (product & chain inside):
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = 2x\ln x + x^{2}\cdot \dfrac{1}{x}].
   [ = 2x\ln x + x ].
3. Multiply by [y]:
   [\dfrac{dy}{dx}][ = x^{x^{2}}\big(2x\ln x + x\big)].
4. Factor [x] if desired:
   [ = x^{x^{2}}\cdot x(2\ln x + 1)][ = x^{x^{2}+1}(2\ln x + 1) ] (either form acceptable).

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

Question 10.
Differentiate [y = \dfrac{( \sin x )^{2}}{x^{\sin x}}], domain where [x>0] and [\sin x>0] for power simplicity.

Step-by-Step Solution:

1. Take ln:
   [\ln y][ = \ln\big((\sin x)^{2}\big) – \ln\big(x^{\sin x}\big)][ = 2\ln(\sin x) – \sin x\ln x].
2. Differentiate implicitly:
   [\dfrac{1}{y}\dfrac{dy}{dx}][ = 2\cdot \dfrac{\cos x}{\sin x} – \big(\cos x\ln x + \sin x\cdot \dfrac{1}{x}\big)]
   (product rule on [\sin x\ln x] with derivative [\cos x\ln x + \sin x\cdot \dfrac{1}{x}]).
3. Simplify RHS:
   [= 2\cot x – \cos x\ln x – \dfrac{\sin x}{x}].
4. Multiply by [y]:
   [\dfrac{dy}{dx}][ = \dfrac{(\sin x)^{2}}{x^{\sin x}}\Big(2\cot x – \cos x\ln x – \dfrac{\sin x}{x}\Big)].
5. (Optional) Replace [\cot x = \dfrac{\cos x}{\sin x}] if a common denominator is desired.

Conclusion:
[\dfrac{d}{dx}\Big(\dfrac{(\sin x)^{2}}{x^{\sin x}}\Big)][ = \dfrac{(\sin x)^{2}}{x^{\sin x}}\Big(2\cot x – \cos x\ln x – \dfrac{\sin x}{x}\Big)], valid under stated domain conditions.