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

Continuity & Differentiability

Q.2. Differentiate the following w.r.t. x : (x)^cosx + (cosx)^x.

Solution :

Let y = (x)cos^x + (cosx)^x
= u + v,
where, u = (x)cosx and v = (cosx)^x.
Therefore, dy/dx = du/dx + dv/dx ----------------------- (1)
Now, u = (x)cosx => logu = cosx logx
Then 1/u du/dx = cosx/x – sinx logx
Or, du/dx = (x)cosx[cosx/x – sinx logx] ------------ (2)
And v = (cosx)^x => log v = x logcosx
Then 1/v dv/dx = 1. logcosx + x. 1.cosx (– sinx) = logcosx – x tanx
Or, dv/dx = (cosx)x[logcosx – x tanx] ----------------- (3)
Using (1), (2) and (3), we get
dy/dx = (x)cosx[cosx/x – sinx logx] + (cosx)x[logcosx –x tanx]. [Ans.]

Q.3. If y = x^cotx + (sin x)^x, find dy/dx.

Solution :

Do yourself. [Ans. dy/dx = x^cotx[(cos x)/x – cosec² xlogx] + (sin x)^x[x cot x + log sin x]

Q.4. If y = (logx)cos x + (x² + 1)/(x² – 1), find dy/dx.

Solution :

We have, y = (logx)^cosx + (x² + 1)/(x² – 1) = u + v
Where u = (logx)^cosx and v = (x² + 1)/(x² – 1)
Therefore, dy/dx = du/dx + dv/dx .
Let u = (logx)^cosx
Taking log of both sides, we get
logu = log[(logx)cos x] = cos x log(logx)
Differentiatingw.r.t.x, we get
1/udu/dx = cos x.(1/logx).1/x + log(logx)(– sin x)
Or, du/dx = u[cos x/(xlogx) – sin xlog(logx)]
= (logx)cos x[cos x/(x logx) – sin x log(logx)]
Again, taking v = (x² + 1)/(x² – 1)
Differentiating w.r.t.x, we get
dv/dx = [(x² – 1).2x – (x² + 1).2x]/(x² – 1)²
= (2x³ – 2x – 2x³ – 2x)/(x² – 1)²
= – 4x/(x² – 1)².
Therefore, dy/dx = (logx)^cosx[cos x/(xlogx) – sin x log(logx)] – 4x/(x² – 1)². [Ans.]

5.6. Derivatives of Functions in Parametric Forms.

Q.1. If x = a(cos t + log tan t/2) and y = a sin t, find dy/dx.

Solution :

We have, x = a(cos t + log tan t/2); y = a sin t.
dx/dt = a [– sin t + {1/(tan t/2)}.(sec2 t/2).(1/2)]
= a [– sin t + {(cos t/2)/(sin t/2)}.{1/cos2 t/2}(1/2)]
= a [– sin t + 1/(2sin t/2 cos t/2)]
= a [– sin t + 1/sin t]
= a [(– sin2 t + 1)/sin t]
= acos2 t/sin t -------------------------- (1)
dy/dt = a cos t -------------------------- (2)
Therefore, dy/dx = (dy/dt)/(dx/dt)
= (acos² t/sin t)/(a cos t)
= (acos² t/sin t)(1/acos t)
= cos t/sin t
= cot t. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289