Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

(I) Differentiate the following w.r.t. x

Question 1.

y = \(x^{\frac{4}{3}}+e^{x}-\sin x\)

Solution:

Question 2.

y = √x + tan x – x^{3}

Solution:

Question 3.

y = log x – cosec x + \(5^{x}-\frac{3}{x^{\frac{3}{2}}}\)

Solution:

Question 4.

y = \(x^{\frac{7}{3}}+5 x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}}\)

Solution:

Question 5.

y = 7^{x} + x^{7} – \(\frac{2}{3}\) x√x – log x + 7^{7}

Solution:

Question 6.

y = 3 cot x – 5e^{x} + 3 log x – \(\frac{4}{x^{\frac{3}{4}}}\)

Solution:

(II) Diffrentiate the following w.r.t. x

Question 1.

y = x^{5} tan x

Solution:

Question 2.

y = x^{3} log x

Solution:

Question 3.

y = (x^{2} + 2)^{2} sin x

Solution:

Question 4.

y = e^{x} log x

Solution:

Question 5.

y = \(x^{\frac{3}{2}} e^{x} \log x\)

Solution:

Question 6.

y = \(\log e^{x^{3}} \log x^{3}\)

Solution:

(III) Diffrentiate the following w.r.t. x

Question 1.

y = x^{2}√x + x^{4} log x

Solution:

Question 2.

y = e^{x} sec x – \(x^{\frac{5}{3}}\) log x

Solution:

Question 3.

y = x^{4} + x√x cos x – x^{2} e^{x}

Solution:

Question 4.

y = (x^{3} – 2) tan x – x cos x + 7^{x} . x^{7}

Solution:

Question 5.

y = sin x log x + e^{x} cos x – e^{x} √x

Solution:

Question 6.

y = e^{x} tan x + cos x log x – √x 5^{x}

Solution:

(IV) Diffrentiate the following w.r.t.x.

Question 1.

y = \(\frac{x^{2}+3}{x^{2}-5}\)

Solution:

Question 2.

y = \(\frac{\sqrt{x}+5}{\sqrt{x}-5}\)

Solution:

Question 3.

y = \(\frac{x e^{x}}{x+e^{x}}\)

Solution:

Question 4.

y = \(\frac{x \log x}{x+\log x}\)

Solution:

y = \(\frac{x \log x}{x+\log x}\)

Question 5.

y = \(\frac{x^{2} \sin x}{x+\cos x}\)

Solution:

Question 6.

y = \(\frac{5 e^{x}-4}{3 e^{x}-2}\)

Solution:

(V).

Question 1.

If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).

Solution:

Let f(x) = ax^{2} + bx + c …..(i)

∴ f(0) = a(0)^{2} + b(0) + c

∴ f(0) = c

But, f(0) = 3 …..(given)

∴ c = 3 …..(ii)

Differentiating (i) w.r.t. x, we get

f'(x) = 2ax + b

∴ f'(2) = 2a(2) + b

∴ f'(2) = 4a + b

But, f'(2) = 2 …..(given)

∴ 4a + b = 2 …..(iii)

Also, f'(3) = 2a(3) + b

∴ f'(3) = 6a + b

But, f'(3) = 12 …..(given)

∴ 6a + b = 12 …..(iv)

equation (iv) – equation (iii), we get

2a = 10

∴ a = 5

Substituting a = 5 in (iii), we get

4(5) + b = 2

∴ b = -18

∴ a = 5, b = -18, c = 3

∴ f(x) = 5x^{2} – 18x + 3

Check:

If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.

f(x) = 5x^{2} – 18x + 3 and f'(x) = 10x – 18

f(0) = 5(0)^{2} – 18(0) + 3 = 3

f'(2) = 10(2) – 18 = 2

f'(3) = 10(3) – 18 = 12

Thus, our answer is correct.

Question 2.

If f(x) = a sin x – b cos x, f'(\(\frac{\pi}{4}\)) = √2 and f'(\(\frac{\pi}{6}\)) = 2, then find f(x).

Solution:

f(x) = a sin x – b cos x

Differentiating w.r.t. x, we get

f'(x) = a cos x – b (- sin x)

∴ f'(x) = a cos x + b sin x

Now, f(x) = a sin x – b cos x

∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x

VI. Fill in the blanks. (Activity Problems)

Question 1.

y = e^{x} . tan x

Diff. w.r.t. x

Solution:

Question 2.

y = \(\frac{\sin x}{x^{2}+2}\)

diff. w.r.t. x

Solution:

Question 3.

y = (3x^{2} + 5) cos x

Diff. w.r.t. x

Solution:

Question 4.

Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of \(\frac{u}{v}\) and \(\frac{1}{v}\) respectively.

Solution: