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) cot^{n} x

Solution:

\(\frac{dy}{dx}\) = n. cot^{n-1} x. \(\frac{d}{dx}\) (cot x)

= n. cot^{n-1} x (- cosec² x)

= – n. cot^{n-1} x. cosec² x

ii) cosec^{4} x

Solution:

\(\frac{dy}{dx}\) = 4.cosec³ x. \(\frac{d}{dx}\)(cosec x)

= 4. cosec³ x (- cosec x. cot x)

= – 4. cosec^{4} x. cot x

iii) tan (e^{x})

=sec 2(e^{x}).(e^{x})¹

= e^{x}. sec² (e^{x})

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) sin^{m} x cos^{n} x

Solution:

\(\frac{dy}{dx}\) = (sin^{m} x). \(\frac{d}{dx}\) (cos^{n} x) + (cos^{n} x) \(\frac{d}{dx}\) (sin^{m} x)

= sin^{m} xn + cos^{n-1} x(-sin x) + cos^{n} x. m sin^{m-1} x. cos x

= m. cos^{n+1} x. sin^{m-1} x – n. sin^{m+1} x. cos^{n-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} (e^{x}))

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 = e^{sin-1x}

Solution:

xviii) y = cos (log x + e^{x})

Solution:

\(\frac{dy}{dx}\) = -sin(log x + e^{x}) = \(\frac{d}{dx}\) (log x + e^{x})

= -sin (log x + e^{x}) (\(\frac{1}{x}\) + e^{x})

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 = e^{sinh y}

Solution:

\(\frac{dy}{dx}\) = e^{sinh y} \(\frac{d}{dx}\) (sinh y)

= e^{sinh 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 = e^{x}

√y = e^{x} – 1

y = (e^{x} – 1)²

\(\frac{dy}{dx}\) = 2(e^{x} – 1) . e^{x} = 2 √y . e^{x}

= 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} (e^{x})]

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^{-1}x)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 = \(\frac{\cos x}{1+\cos x}\)