250+ TOP MCQs on Logarithmic Differentiation | Class 12 Maths

By /

Mathematics Multiple Choice Questions on “Logarithmic Differentiation”.

1. Differentiate (log⁡2x)sin⁡3x with respect to x.
a) (3 cos⁡3x log⁡(log⁡2x)+(frac{sin⁡3x}{x log⁡2x}))
b) (log⁡2x^{sin⁡3x} ,(3 ,cos⁡3x ,log⁡(log⁡2x)+frac{sin⁡3x}{x ,log⁡2x}))
c) –((3 ,cos⁡3x ,log⁡(log⁡2x)+frac{sin⁡3x}{x log⁡2x}))
d) (frac{3 ,cos⁡3x ,log⁡(log⁡2x)+frac{sin⁡3x}{x log⁡2x}}{log⁡2x^{sin⁡3x}})
Answer: b
Clarification: Consider y=((log⁡2x)^{sin⁡3x})
Applying log on both sides, we get
log⁡y=(log⁡(log⁡2x)^{sin⁡3x})
log⁡y=sin⁡3x log⁡(log⁡2x)
Differentiating with respect to x, we get
(frac{1}{y} ,frac{dy}{dx}=log⁡(log⁡2x)frac{d}{dx} (sin⁡3x)+sin⁡3x frac{d}{dx} ,(log⁡(log⁡2x)))
By using chain rule, we get
(frac{1}{y} ,frac{dy}{dx}=log⁡(log⁡2x).3 ,cos⁡3x+sin⁡3x.frac{1}{log⁡2x}.frac{1}{2x}.2 ,(∵u.v=u’ ,v+uv’))
(frac{dy}{dx})=y(3 cos⁡3x log⁡(log⁡2x)+(frac{sin⁡3x}{x ,log⁡2x}))
∴(frac{dy}{dx})=log⁡2xsin⁡3x (left (3 ,cos⁡3x ,log⁡(log⁡2x)+frac{sin⁡3x}{x ,log⁡2x} right ))

2. Differentiate 4xex with respect to x.
a) xex e-x (x log⁡x+1)
b) -4xex-1 ex (x log⁡x+1)
c) 4xex ex (x log⁡x+1)
d) 4xex-1 ex (x log⁡x+1)
Answer: d
Clarification: Consider y=4xex
Applying log on both sides, we get
log⁡y=log⁡4xex
log⁡y=log⁡4+log⁡xex (∵log⁡ab=log⁡a+log⁡b)
Differentiating both sides with respect to x, we get
(frac{1}{y} frac{dy}{dx}=0+frac{d}{dx}(e^x ,log⁡x)(∵log⁡a^b=b ,log⁡a))
(frac{1}{y} frac{dy}{dx}=frac{d}{dx} ,(e^x) ,log⁡x+e^{x} ,frac{d}{dx} ,(log⁡x))
(frac{dy}{dx}=y(e^x log⁡x+frac{e^x}{x}))
(frac{dy}{dx}=frac{4x^{e^{x}}e^x ,(x log⁡x+1)}{x}=4x^{e^x-1} ,e^x ,(x log⁡x+1)).

3. Differentiate 9tan⁡3x with respect to x.
a) 9tan⁡3x (3 log⁡9 sec2⁡x)
b) 9tan⁡3x (3 log⁡3 sec2⁡⁡x)
c) 9tan⁡3x (3 log⁡9 sec⁡x)
d) -9tan⁡3x (3 log⁡9 sec2⁡⁡x)
Answer: a
Clarification: Consider y=9tan⁡3x
Applying log on both sides, we get
log⁡y=log⁡9tan⁡3x
Differentiating both sides with respect to x, we get
(frac{1}{y} frac{dy}{dx}=frac{d}{dx} )(tan⁡3x.log⁡9)
(frac{1}{y} frac{dy}{dx}=frac{d}{dx} ,(tan⁡3x) ,log⁡9+frac{d}{dx} ,(log⁡9).tan3x ,(∵ Using ,u.v=u’ ,v+uv’))
(frac{dy}{dx})=y(3 sec2⁡⁡x.log⁡9+0)
(frac{dy}{dx})=9tan⁡3x (3 log⁡9 sec2⁡x)

4. Differentiate (cos⁡3x)3x with respect to x.
a) (cos⁡3x)x (3 log⁡(cos⁡3x) – 9x tan⁡3x)
b) (cos⁡3x)3x (3 log⁡(cos⁡3x) + 9x tan⁡3x)
c) (cos⁡3x)3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)
d) (cos⁡3x)3x (log⁡(cos⁡3x) + 9 tan⁡3x)

Answer: c
Clarification: Consider y=(cos⁡3x)3x
Applying log on both sides, we get
log⁡y=log⁡(cos⁡3x)3x
log⁡y=3x log⁡(cos⁡3x)
Differentiating both sides with respect to x, we get
(frac{1}{y} frac{dy}{dx}=frac{d}{dx} (3x ,log⁡(cos⁡3x)))
By using u.v=u’ v+uv’, we get
(frac{1}{y} frac{dy}{dx})=(frac{d}{dx} ,(3x) ,log⁡(cos⁡3x)+frac{d}{dx} ,(log⁡(cos⁡3x)).3x)
(frac{dy}{dx})=y(3 log⁡(cos⁡3x) + (frac{1}{cos⁡3x} ,. frac{d}{dx} ,(cos⁡3x).3x))
(frac{dy}{dx})=y(3 log⁡(cos⁡3x) + (frac{1}{cos⁡3x} ,. ,(-sin⁡3x).frac{d}{dx}(3x).3x))
(frac{dy}{dx})=y(3 log⁡(cos⁡3x) + (frac{1}{cos⁡3x} ,. ,(-sin⁡3x).3.3x))
(frac{dy}{dx})=y(3 log⁡(cos⁡3x) – 9x tan⁡3x)
(frac{dy}{dx})=(cos⁡3x)3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

5. Differentiate 7x(2e2x) with respect to x.
a) 14e2x x(2e2x) (2 log⁡x+(frac{1}{x}))
b) 14x(2e2x) (2 log⁡x+(frac{1}{x}))
c) 14e2x x(2e2x) (2 log⁡x-(frac{1}{x}))
d) 14e2x x(2e2x) (log⁡x-(frac{1}{x}))
Answer: a
Clarification: Consider y=7x(2e2x)
log⁡y=log⁡7x(2e2x)
log⁡y=log⁡7+log⁡x(2e2x)
log⁡y=log⁡7+2e2x log⁡x
Differentiating with respect to x on both sides, we get
(frac{1}{y} frac{dy}{dx})=(frac{d}{dx}) (log⁡7+2e2x log⁡x)
(frac{1}{y} frac{dy}{dx})=0+(frac{d}{dx}) (2e2x) log⁡x+(frac{d}{dx}) (log⁡x)2e2x (using u.v=u’ v+uv’)
(frac{1}{y} frac{dy}{dx})=2e2x.2.log⁡x+(frac{2e^{2x}}{x})
(frac{dy}{dx})=y( left (4e^{2x} ,log⁡x+frac{2e^{2x}}{x}right))
(frac{dy}{dx})=7x(2e2x) ( left (4e^{2x} ,log⁡x+frac{2e^{2x}}{x}right))
(frac{dy}{dx})=14e2x x(2e2x) (2 log⁡x+(frac{1}{x}))

6. Differentiate (e^{4x^5}.2x^{log⁡x^2}) with respect to x.
a) (e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2))
b) (4e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2))
c) (4e^{4x^5}.x^{log⁡x^2-1} (10x^5-log⁡2x^2))
d) (x^{log⁡x^2 -1} (10x^4+log⁡2x^2))
Answer: b
Clarification: Consider y=(e^{4x^5}+2x^{log⁡x^2})
Applying log on both sides, we get
log⁡y=(log⁡e^{4x^5} ,+ ,log⁡2x^{log⁡x^2})
log⁡y=(4x^5+log⁡x^2 ,. ,log⁡2x)
log⁡y=(4x^5+2 ,log⁡x ,log⁡2x)
Differentiating with respect to x, we get
(frac{1}{y} frac{dy}{dx})=(20x^4+2(frac{d}{dx} ,(log⁡x) ,log⁡2x+frac{d}{dx} ,(log⁡2x) ,log⁡x))
(frac{1}{y} frac{dy}{dx})=(20x^4+2left (frac{log⁡2x}{x}+frac{1}{2x}.2.log⁡xright ))
(frac{1}{y} frac{dy}{dx})=(20x^4+frac{2(log⁡2x+log⁡x)}{x})
(frac{1}{y} frac{dy}{dx})=(20x^4+frac{2(log⁡2x^2)}{x})
(frac{dy}{dx})=(y(20x^4+frac{2(log⁡2x^2)}{x}))
(frac{dy}{dx})=(e^{4x^5}.2x^{log⁡x^2} (20x^4+frac{2(log⁡2x^2)}{x}))
(frac{dy}{dx})=(4e^{4x^5}.x^{log⁡x^2 -1} (10x^5+log⁡2x^2))

7. Differentiate 2(tan⁡x)cot⁡x with respect to x.
a) 2 csc2⁡x.tan⁡xcot⁡x (1-log⁡(tan⁡x))
b) csc2⁡x.tan⁡xcot⁡x (1-log⁡(tan⁡x))
c) 2 csc2⁡x.tan⁡xcot⁡x (1+log⁡(tan⁡x))
d) 2tan⁡xcot⁡x (1-log⁡(tan⁡x))

Answer: a
Clarification: Consider y=2(tan⁡x)cot⁡x
Applying log in both sides,
log⁡y=log⁡2(tan⁡x)cot⁡x
log⁡y=log⁡2+log⁡(tan⁡x)cot⁡x
log⁡y=log⁡2+cot⁡x log⁡(tan⁡x)
Differentiating both sides with respect to x, we get
(frac{1}{y} frac{dy}{dx}=0+frac{d}{dx} ,(cot⁡x) ,log⁡(tan⁡x)+cot⁡x frac{d}{dx} ,(log⁡(tan⁡x)))
(frac{1}{y} frac{dy}{dx}=-csc^{2⁡}x.log⁡(tan⁡x)+cot⁡x.frac{1}{tan⁡x}.sec^{2⁡}x)
(frac{dy}{dx} = yleft(-csc^{2⁡x}.log⁡(tan⁡x)+frac{(1+tan^{2⁡x})}{tan^{2⁡x}}right))
(frac{dy}{dx})=2(tan⁡x)cot⁡x (left (-csc^{2⁡x} log⁡(tan⁡x)+cot^{2⁡x}+1 right ))
(frac{dy}{dx})=2(tan⁡x)cot⁡x ((-csc^{2⁡x} log⁡(tan⁡x)+csc^{2⁡x}))
(frac{dy}{dx})=2(tan⁡x)cot⁡x (csc2⁡x (1-log⁡(tan⁡x))
∴(frac{dy}{dx})=2 csc2⁡x.tan⁡xcot⁡x (1-log⁡(tan⁡x))

8. Differentiate (3 cos⁡x)x with respect to x.
a) (3 cos⁡x)x (log⁡(3 cos⁡x)+x tan⁡x)
b) (3 cos⁡x)x (log⁡(3 cos⁡x)+tan⁡x)
c) (cos⁡x)^x (log⁡(3 cos⁡x)-x tan⁡x)
d) (3 cos⁡x)x (log⁡(3 cos⁡x)-x tan⁡x)
Answer: d
Clarification: Consider y=(3 cos⁡x)x
Applying log on both sides, we get
log⁡y=log⁡(3 cos⁡x)x
log⁡y=x log⁡(3 cos⁡x)
log⁡y=x(log⁡3+log⁡(cos⁡x))
Differentiating both sides with respect to x, we get
(frac{1}{y} frac{dy}{dx} =frac{d}{dx} (x log⁡3)+frac{d}{dx} (x) log⁡(cos⁡x)+frac{d}{dx}(log⁡(cos⁡x)).x)
(frac{1}{y} frac{dy}{dx}=log⁡3+log⁡(cos⁡x)+frac{1}{cos⁡x}.-sin⁡x.x)
(frac{1}{y} frac{dy}{dx})=log⁡3+log⁡(cos⁡x)-x tan⁡x
(frac{dy}{dx})=y(log⁡(3 cos⁡x)-x tan⁡x)
(frac{dy}{dx})=(3 cos⁡x)x (log⁡(3 cos⁡x)-x tan⁡x)

9. Differentiate (sqrt{frac{x+1}{3x-1}}) with respect to x.
a) (frac{-2}{(3x-1)sqrt{(3x-1)(x+1)}})
b) (frac{2}{(3x-1)sqrt{(3x-1)(x+1)}})
c) (frac{1}{(3x-1)sqrt{(3x-1)(x+1)}})
d) (frac{-2}{sqrt{(3x-1)(x+1)}})
Answer: a
Clarification: Consider y=(sqrt{frac{x+1}{3x-1}})
Applying log to both sides, we get
log⁡y=log⁡(sqrt{frac{x+1}{3x-1}})
log⁡y=(frac{1}{2} log⁡left (frac{x+1}{3x-1}right ))
log⁡y=(frac{1}{2}) (log⁡(x+1)-log⁡(3x-1))
Differentiating with respect to x, we get
(frac{1}{y} frac{dy}{dx})=(frac{1}{2}left (frac{d}{dx} (log⁡(x+1))-frac{d}{dx} (log⁡(3x-1))right ))
(frac{1}{y} frac{dy}{dx})=(frac{1}{2}left (frac{1}{x+1}-frac{3}{3x-1}right ))
(frac{1}{y} frac{dy}{dx})=(frac{1}{2}left (frac{3x-1-3x-3}{(x+1)(3x-1)})right ))
(frac{1}{y} frac{dy}{dx})=(frac{1}{2}left (frac{-4}{(x+1)(3x-1)}right ))
(frac{dy}{dx})=(sqrt{frac{x+1}{3x-1}} left (frac{-2}{(x+1)(3x-1)}right ))
(frac{dy}{dx})=(frac{-2}{(3x-1) sqrt{(3x-1)(x+1)}})

10. Differentiate x3ex with respect to x.
a) 3e3x (3 log⁡x+(frac{1}{x}))
b) x3e3x.3e3x (3 log⁡x-(frac{1}{x}))
c) x3e3x (3 log⁡x+(frac{1}{x}))
d) x3e3x.3e3x (3 log⁡x+(frac{1}{x}))
Answer: d
Clarification: Consider y=x3e3x
Applying log on both sides, we get
log⁡y=3e3x log⁡x
Differentiating both sides with respect to x, we get
(frac{1}{y} frac{dy}{dx} = frac{d}{dx} (3e^{3x}) log⁡x+frac{d}{dx} (log⁡x)3e^{3x})
(frac{1}{y} frac{dy}{dx})=3e3x.3.log⁡x+(frac{1}{x}) 3e3x
(frac{dy}{dx})=y(3e3x.3.log⁡x+(frac{1}{x}) 3e3x)
(frac{dy}{dx})=x3e3x.3e3x (3 log⁡x+(frac{1}{x}))

  250+ TOP MCQs on Determinants – Adjoint and Inverse of a Matrix | Class 12 Maths

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top