Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Differentiation Solutions Exercise 9(b) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Differentiation Solutions Exercise 9(b)

I. Compute the following limits.

Question 1.
Find the derivatives of the following function.
i) cotn x
Solution:
$$\frac{dy}{dx}$$ = n. cotn-1 x. $$\frac{d}{dx}$$ (cot x)
= n. cotn-1 x (- cosec² x)
= – n. cotn-1 x. cosec² x

ii) cosec4 x
Solution:
$$\frac{dy}{dx}$$ = 4.cosec³ x. $$\frac{d}{dx}$$(cosec x)
= 4. cosec³ x (- cosec x. cot x)
= – 4. cosec4 x. cot x

iii) tan (ex)
=sec 2(ex).(ex
= ex. sec² (ex)

iv) $$\frac{1-\cos 2 x}{1+\cos 2 x}$$
Solution:
$$\frac{2 \sin ^{2} x}{2 \cos ^{2} x}$$ = tan² x $$\frac{dy}{dx}$$ = 2 tan x . sec² x

v) sinm x cosn x
Solution:
$$\frac{dy}{dx}$$ = (sinm x). $$\frac{d}{dx}$$ (cosn x) + (cosn x) $$\frac{d}{dx}$$ (sinm x)
= sinm xn + cosn-1 x(-sin x) + cosn x. m sinm-1 x. cos x
= m. cosn+1 x. sinm-1 x – n. sinm+1 x. cosn-1 x.

vi) sin mx. cos nx
Solution:
$$\frac{dy}{dx}$$ = sin mx $$\frac{d}{dx}$$ (cos nx) + (cos nx) $$\frac{dy}{dx}$$ (sin mx)
= sin mx (-n sin nx)+cos nx (m cos mx)
= m. cos mx. cos nx – n. sin mx . sin nx

vii) x tan-1 x
Solution:
$$\frac{dy}{dx}$$ = x. $$\frac{d}{dx}$$ (tan-1 x) + (tan-1 x) $$\frac{d}{dx}$$ (x)
= $$\frac{x}{1+x^{2}}$$ + tan-1 x.

viii) sin-1 (cos x)
Solution:
= sin-1 [sin ($$\frac{\pi}{2}$$ – x)] = $$\frac{\pi}{2}$$ – x
$$\frac{dy}{dx}$$ = 0 – 1 = -1

ix) log (tan 5x)
Solution:

x) sinh-1 ($$\frac{3x}{4}$$)
Solution:

xi) tan-1 (log x)
Solution:

xii) log ($$\frac{x^{2}+x+2}{x^{2}-x+2}$$)
Solution:
$$\frac{dy}{dx}$$ = log(x² + x + 2) – log(x² – x + 2)

xiii) log (sin-1 (ex))
Solution:

xiv) (sin x)² (sin-1 x)²
Solution:

xv) y = $$\frac{\cos x}{\sin x+\cos x}$$
Solution:

xvi) $$\frac{x\left(1+x^{2}\right)}{\sqrt{1-x^{2}}}$$
Solution:

xvii) y = esin-1x
Solution:

xviii) y = cos (log x + ex)
Solution:
$$\frac{dy}{dx}$$ = -sin(log x + ex) = $$\frac{d}{dx}$$ (log x + ex)
= -sin (log x + ex) ($$\frac{1}{x}$$ + ex)

xix) y = $$\frac{\sin (x+a)}{\cos x}$$
Solution:

xx) y = cot-1 (coses 3x)
Solution:

Question 2.
Find the derivatives of the following fountion.
i) x = sinh² y
Solution:
$$\frac{dy}{dx}$$ = 2 sinh y . cosh y

ii) x = tanh² y
Solution:
$$\frac{dy}{dx}$$ = 2 tanh y . sech² y

iii) x = esinh y
Solution:
$$\frac{dy}{dx}$$ = esinh y $$\frac{d}{dx}$$ (sinh y)
= esinh y . cosh y
= x. cosh y
$$\frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{x \cdot \cosh y}$$

iv) x =tan (e-y)
Solution:
$$\frac{dy}{dx}$$ = sec² (e-y) . (e-y)¹ = -e-y . sec² e-y
= -e-y(1 + tan² (e-y) = – e-y(1 + x²)
$$\frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=-\frac{1}{e^{-y}\left(1+x^{2}\right)}=-\frac{e^{y}}{1+x^{2}}$$

v) x = log (1 + sin² y)
Solution:

vi) x = log (1 + √y)
Solution:
1 + √y = ex
√y = ex – 1
y = (ex – 1)²
$$\frac{dy}{dx}$$ = 2(ex – 1) . ex = 2 √y . ex
= 2 √y (√y + 1)
= 2(y + √y)

II. Find the derivativies of the following functions.

i) y = cos [log (cot x)]
Solution:

ii) y = sin-1 $$\frac{1-x}{1+x}$$
Solution:

iii) log (cot (1 – x²))
Solution:

iv) y = sin [cos (x²)]
Solution:
$$\frac{dy}{dx}$$ = cos [cos (x²)].$$\frac{d}{dx}$$[cos (x²)]
= cos [cos (x²)](sin (x²)).$$\frac{d}{dx}$$(x²)
= – 2x. sin (x²). cos [cos (x²)]

v) y = sin [tan-1 (ex)]
Solution:

vi) y = $$\frac{\sin (a x+b)}{\cos (c x+d)}$$
Solution:

vii) y = tan-1 (tan h $$\frac{x}{2}$$)
Solution:

viii) y = sinx . (Tan-1x)²
Solution:

III. Find the derivatives of the following functions.

Question 1.
y = sin-1 $$\left(\frac{b+a \sin x}{a+b \sin x}\right)$$ (a > 0, b > 0)
Solution:
Let u = $$\frac{b+a \sin x}{a+b \sin x}$$

Question 2.
cos-1$$\left(\frac{b+a \cos x}{a+b \cos x}\right)$$ (a > 0, b > 0)
Solution:
Let u = $$\frac{b+a \cos x}{a+b \cos x}$$

Question 3.
tan-1 $$\left[\frac{\cos x}{1+\cos x}\right]$$
Solution:
Let u = $$\frac{\cos x}{1+\cos x}$$