Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.4

Find the derivatives of the following:

Question 1.
y = x^{cos x}
Answer:
y = x^{cos x}
Taking log on both sides
log y = log x^{cos x}
log y = cos x log x
Differentiating with respect to x

Question 2.
y = x^{log x} + (log x)^x
Answer:
y = x^{log x} + (log x)^x
Let u = x^{log x}, v = (log x)^x
log u = log x^{log x}
log u = (log x) (log x)
log u = (log x)²

v = (log x)^x
log v = log (log x)^x
log v = x log (log x)

Question 3.
\(\sqrt{x y}\) = e^{(x – y)}
Answer:

Question 4.
x^y = y^x
Answer:
x^y = y^x
Taking log on both sides
log x^y = log y^x
y log x = x log y
Differentiate with respect to x

Question 5.
(cos x)^{log x}
Answer:
y = (cos x)^{log x}
Taking log on both sides
log y = log (cos x)^{log x}
log y = (log x) log (cos x)
Differentiating with respect to x

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:

Question 7.
\(\sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)\)
Answer:

Question 8.
tan (x + y) + tan (x – y) = x
Answer:
tan (x + y) + tan (x – y) = x
Differentiating with respect to x

Question 9.
If cos(xy) = x, show that
\(\frac{d y}{d x}=\frac{-(1+y \sin (x y))}{x \sin x y}\)
Answer:
cos (xy) = x
Differentiating with respect to x

Question 10.
\(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Answer:
Let y = \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
[1 – cos 2θ = 2 sin²θ and 1 + cos 2θ = 2 sin² θ]

Question 11.
tan^-1 = \(\left(\frac{6 x}{1-9 x^{2}}\right)\)
Answer:

Question 12.
\(\cos \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\)
Answer:

Question 13.
x = a cos^t ; y = a sin³t
Answer:
x = a cos^t , y = a sin³t

Question 14.
x = a (cos t + t sin t);
y = a (sin t – t cos t)
Answer:
x = a (cos t + t sin t) , y = a (sin t – t cos t)
\(\frac{d x}{d t}\) = a [- sin t + t cos t + sin t ]
\(\frac{d x}{d t}\) = at cos t —— (1)
y = a (sin t – t cos t)
\(\frac{d x}{d t}\) = a [cos t – (t × – sin t + cos t × 1)]
\(\frac{d x}{d t}\) = a[cos t + t sin t – cos t]
\(\frac{d x}{d t}\) = at sin t —— (2)
From equations (1) and (2) we get

Question 15.
x = \(\frac{1-t^{2}}{1+t^{2}}\) , y = \(\frac{2 t}{1+t^{2}}\)
Answer:

Question 16.
cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Answer:
Let y = cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Put x = tan θ

y = cos^-1 (cos 2θ)
y = 2θ
y = 2 tan^-1 x

Question 17.
sin^-1 (3x – 4x³)
Answer:
Let y = sin^-1 (3x – 4x³)
Put x = sin θ
y = sin^-1 (3 sin θ – 4 sin³ θ)
y = sin^-1 (sin 3θ)
y = 3θ
y = 3 sin^-1 x
\(\frac{d y}{d x}=\frac{3}{\sqrt{1-x^{2}}}\)

Question 18.
tan^-1 \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\)
Answer:
Let y = tan^-1 \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\)

Question 19.
Find the derivative of sin x² with respect to x².
Answer:
Let u = sin x²
\(\frac{\mathrm{d} \mathrm{u}}{\mathrm{d} x}\) = cos (x²) × 2x
\(\frac{\mathrm{d} \mathrm{u}}{\mathrm{d} x}\) = 2x cos (x²)
Let v = x²

Question 20.
Find the derivative of sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\) with respect to tan^-1 x.
Answer:

Question 21.
If u = tan^-1\(\frac{\sqrt{1+x^{2}}-1}{x}\) and v = tan^-1x, find \(\frac{\mathrm{du}}{\mathrm{dv}}\)
Answer:

Question 22.
Find the derivative with tan^-1\(\left(\frac{\sin x}{1+\cos x}\right)\) with respect to tan^-1\(\left(\frac{\cos x}{1+\sin x}\right)\)
Answer:

Question 23.
If y = sin^-1x then find y”.
Answer:
y = sin^-1 x

Question 24.
If y = e^tan-1x, show that (1 + x²) y” + (2x – 1) y’ = 0
Answer:
y = e^tan-1x
y = e^tan-1x \(\left(\frac{1}{1+x^{2}}\right)\)
⇒ y’ = \(\frac{y}{1+x^{2}}\) ⇒ y'(1 + x²) = y
differentiating w.r.to x
y’ (2x) + (1 + x²) (y”) = y’
(i.e.) (1 + x²) y” + y’ (2x) – y’ = 0
(i.e.) (1 + x²) y” + (2x – 1) y’ = 0

Question 25.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\), show that (1 – x²)y₂ – 3xy₁ – y = 0.
Answer:

Question 26.
If x = a (θ + sin θ), y = a (1 – cos θ) then prove that at θ = \(\frac{\pi}{2}\), y” = \(\frac{1}{a}\)
Answer:

Question 27.
If sin y = x sin (a + y), the prove that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\), a ≠ nπ
Answer:
Given sin y = x sin (a + y) ——- (1)
Differentiating with respect to x , we get
cos y \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = x cos (a + y) (0 + \(\frac{\mathrm{dy}}{\mathrm{d} x}\)) + sin (a + y) . 1

Question 28.
If y = (cos^-1 x)², prove that (1 – x²) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}\) – x \(\frac{\mathrm{dy}}{\mathrm{d} x}\) – 2 = 0. Hence find y₂ when x = 0.
Answer: