CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Continuity & Differentiability

Q.4. If y = sin –1[{5x + 12√(1 – x2)}/13], find dy/dx.

Solution :

We have, y = sin ^-1[{5x + 12 √(1 – x2)}/13]
Let x = sin z => z = sin –1x
Therefore, y = sin –1[{5 sin z + 12√(1 – sin2 z)}/13]
= sin ^-1[(5 sin z + 12 cos z)/13]
= sin ^-1[5/13 sin z + 12/13 cos z]
Let cos t = 5/13 => t = cos ^-1(5/13)
Therefore, sin t = √(1 – cos2 t) = √(1 – 25/169) = √(144/169) = 12/13.
Therefore, y = sin –1[cos t sin z + sin t cos z]
= sin ^-1sin (t + z)
= t + z
= cos ^-1(5/13) + sin ^-1x.
Differentiating w.r.t. x, we get
dy/dx = 0 + 1/√(1 – x²)
= 1/√(1 – x²). [Ans.]

5.4. Derivatives of Exponential and logarithmic Functions.

Q.1. If y = √(x² + 1) – log [1/x + √(1 + 1/x²)], find dy/dx.

Solution :

We have, y = √(x² + 1) – log[1/x + √(1/x + 1/x²)]
= √(x² + 1) – log[1/x + √{(x² + 1)/x²}]
= √(x² + 1) – log[1/x + {√(x² + 1)}/x]
= √(x² + 1) – log[{1 + √(x² + 1)}/x]
= √(x² + 1) – log[1 + √(x² + 1)] + log x
Differentiating w.r.t. x, we get
dy/dx = 1/2(x2 + 1) – (1/2).2x – 1/[1 + √(x² + 1)][0 + 1/2(x² + 1) – (1/2).2x] + 1/x
= x/√(x² + 1) – 1/[1 + √(x² + 1)].[x/√(x² + 1)] + 1/x.
= x/√(x² + 1)[1 – 1/{1 + √(x² + 1)}] + 1/x
= x/√(x² + 1)[{1 + √(x² + 1) – 1}/{1 + √(x² + 1)}] + 1/x
= [x/√(x² + 1)].[√(x² + 1)/{1 + √(x² + 1)}] + 1/x
= x/{1 + √(x² + 1)} + 1/x. [Ans.]

5.5. Logarithmic Differentiation.

Q.1. If y = √[{(x – 3)(x² + 4)}/(3x² + 4x + 5)], find dy/dx.

Solution :

We have, y = √[{(x – 3)(x² + 4)}/(3x² + 4x + 5)]
= [√(x – 3).√(x² + 4)]/√(3x² + 4x + 5)
Taking log of both sides, we get
log y = 1/2log(x – 3) + 1/2log(x² + 4) – 1/2log(3x² + 4x + 5)
Differentiating with respect to x, we get
1/y.dy/dx = 1/2.1/(x – 3).1 + 1/2.1/(x² + 4).2x – 1/2.1/(3x² + 4x + 5).(6x + 4)
= 1/2.1/(x – 3) + x/(x² + 4) – (3x + 2)/(3x² + 4x + 5)
Therefore,
dy/dx = √[{(x – 3)(x² + 4)}/(3x² + 4x + 5)]×[1/{2(x – 3)} + x/(x² + 4) – (3x + 2)/(3x² + 4x + 5)] [Ans.]

Paper By Mr. M.P.Keshari
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