Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Find the derivatives of the following functions with respect to corresponding independent variables:

Question 1.
f(x) = x – 3 sin x
Answer:
f(x) = x – 3 sinx
= f'(x) = 1 – 3 (cos x)
= 1 – 3 cos x

Question 2.
y = sin x + cos x
Answer:
y = sin x + cos x
\(\frac{d y}{d x}\) = cos x – sin x

Question 3.
f(x) = x sin x
Answer:
f(x) = uv
⇒ f'(x) = uv’ + vu’ = u\(\frac{d u}{d x}\) + v\(\frac{d v}{d x}\)
Now u = x ⇒ u’ = 1
v = sin x ⇒ v’ cos x
f'(x) = x (cos x) + sin x(1)
= x cos x + sin x

Question 4.
y = cos x – 2 tan x
Answer:
y = cos x – 2 tan x
\(\frac{d y}{d x}\) = – sin x – 2 sec² x

Question 5.
g(t) = t³ cos t
Answer:
g(t) = t³ cost (i.e.) u = t³ and v = cos t
let u’ = \(\frac{d u}{d x}\) and v’= \(\frac{d v}{d x}\) = (-sint)
g'(t) = uv’ + vu’
g'(t) = t³ (-sin t) + cos t (3t²)
= -t³ sin t + 3t² cos t

Question 6.
g(t) = 4 sec t + tan t
Answer:
g(t) = 4 sec t + tan t
g'(t) = 4 sec t tan t + sec²t

Question 7.
y = e^x sin x
Answer:
y = e^x sin x
⇒ y = uv’ + vu’
Now u = e^x ⇒ u’ = \(\frac{d u}{d x}\) e^x
v = sin x ⇒ v’ = \(\frac{d v}{d x}\) cos x
i.e. y’ = e^x (cos x) + sin x (e^x)
= e^x [sin x + cos x]

Question 8.
y = \(\frac{\tan x}{x}\)
Answer:

Question 9.
y = \(\frac{\sin x}{1+\cos x}\)
Answer:

Question 10.
y = \(\frac{x}{\sin x+\cos x}\)
Answer:

Question 11.
y = \(\frac{\tan x-1}{\sec x}\)
Answer:

Question 12.
y = \(\frac{\sin x}{x^{2}}\)
Answer:

Question 13.
y = tan θ (sin θ + cos θ)
Answer:
y = tan θ (sin θ + cos θ)
\(\frac{d y}{d x}\) = tan θ (cos θ – sin θ) + (sin θ + cos 0) sec² θ
= tan θ cos θ – tan θ sin θ + sin θ sec²θ + cos θ sec²θ

= sin θ – sin²θ sec θ + tan θ sec θ + sec θ
= sin θ + (1 – sin²θ) sec θ + sec θ tan θ
= sin θ + cos²θ × \(\frac{1}{\cos \theta}\) + sec θ tan θ
= sin θ + cos θ + sec θ tan θ

Question 14.
y = cosec x . cot x
Answer:
y = cosec x . cot x
\(\frac{d y}{d x}\) = cosec x × – cosec² x + cot x × – cosec x cot x
= – cosec³ x – cosec x cot² x
= – cosec x (cosec² x + cot² x)

Question 15.
y = x sin x cos x
Answer:

Question 16.
y = e^-x . log x
Answer:
y = e^-x . log x
\(\frac{d y}{d x}\) = e^-x × \(\frac{1}{x}\) + (log x) e^-x (-1)

Question 17.
y = (x² + 5) log (1 + x) e^-3x
Answer:

Question 18.
y = sin x⁰
Answer:

Question 19.
y = log₁₀ x
Answer:
y = log₁₀ x
y = log_ex . log₁₀e

Question 20.
Draw the function f'(x) if f(x) = 2x² – 5x + 3
Answer:
f(x) = 2x² – 5x + 3
f'(x) = 4x – 5 which is a linear function
(i.e.) y = 4x – 5