RBSE Solutions for Class 12 Maths Chapter 7 Differentiation Miscellaneous is part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Miscellaneous.

## Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Miscellaneous Exercise

**Differentiate given functions from question 1 to 10 w.r.t. x :**

Question 1.

sin^{-1} (x\(\sqrt { x }\)), 0 ≤ x ≤ 1.

Solution:

Let y = sin^{-1} (x\(\sqrt { x }\))

Diff. w.r.t. x on both sides,

Question 2.

Solution:

Question 3.

Solution:

Question 4.

x^{3},e^{x},sin x.

Solution:

Let y = x^{3}.e^{x}.sin x

Diff. w.r.t. x on both sides,

Question 5.

log (\(\frac { x }{ { a }^{ x } } \))

Solution:

Question 6.

(x log x)^{log x,}

Solution:

Let y = (x log x)^{log x}

Taking log of both sides,

log y = log (x log x)^{log x}

⇒ log y = log x.log (x log x)

Diff. w.r.t. x on both sides,

Question 7.

Solution:

Question 8.

x^{x2} – 3 + (x – 3)^{x2}, x > 3

Solution:

Question 9.

y – 12(1 – cos t),x = 10(t – sin t).

Solution:

Diff. w.r.t. x on both sides,

Question 10.

sin^{-1} x + sin^{-1} \(\sqrt { 1-{ x }^{ 2 } } \)

Solution:

Question 11.

If cos^{-1}

a, then prove that \(\frac { dy }{ dx } \) = \(\frac { y }{ x } \)

Solution:

Question 12.

If, y = x sin (a + y) then prove

Solution:

Question 13.

If y = (sin x – cos x)^{(sin x- cos x) }then find \(\frac { dy }{ dx } \)

Solution:

Let y – (sin x – cos x)(sin x – cos)

Taking log on both sides,

log y = (sin x – cos x) log (sin x – cos x)

Diff. w.r.t. x on both sides,

Question 14.

If y = sin (sin x), then show that:

Solution:

Question 15.

(i) If y = e^{ax} sin bx, then show that

Solution:

Question 16.

Prove the Rolle’s theorem for the following :

Solution:

which is defined and exist at every point in interval (0, 2).

So, function is differentiable in interval (0, 2).

∵ f(0) = 0 = f(2)

⇒ f(0) = f(2)

Here,f(x) satisfies Rolle’s theorem in given interval.

(b) Given f(x) = (x – 1) (x – 3), x ∈ [1, 3]

Here, f (x) is continuous in interval [1, 3] and f'(x) = 2x – 4, which is defined and exist at every point in interval (1, 3). So,f(x) is differentiable in interval (1, 3).

∵ f(1) = 0 = f(3)

⇒ f(1) = f(3)

Here,f(x) satisfies Rolle’s theorem in given interval.

Hence f'(c) = 0

f'(c) = 2c – 4 = 0

⇒ 2c = 4

⇒ c = 2

∴ c = 2 ∈ (1,3)

such that

f'(c) = 0

Hence, Rolle’s theorem verified for c = 2.

Question 17.

Verfiy the Langrange’s theorem for the following:

Solution:

(a) Given function

f(x) = (x – 1) (x – 2) (x – 3), x ∈ [0, 4]

⇒ f(x) = x^{3} – 6x^{2} + 11x – 6, x ∈ [0, 4]

Clearly,f(x) is continuous in intemval [0,4] and f'(x) is defined and exists in interval (0,4). Hence, function satisfies Lagrange’s mean value theorem.

Hence,f(x) is continuous and differentiable in interval (1, 3) excepting x = 2.

So, function f(x) is not differentiable at x = 2.

Here condition for Lagrange’s theorem does not satisfied. Hence, Lagrange’s does not verified.

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