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

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

Find the derivatives of the following functions
Question 1.
y = x^{cos x}
Solution:
y = x^{cos x}
Taking log on both sides
log y = log x^{cos x} = cos x log x
differentiating w.r.to x we get

Question 2.
y = x^logx + (logx)^x
Solution:
y = x^logx + (logx)^x
Let y = u + v
Then \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
u = x^logx
Taking log on both sides
log u = log x log x = log (x)²
differentiating w.r.to x

Taking log on both sides
log u = log (logx)^x = x log (log x)
differentiating w.r.to x

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

Question 4.
x^y = y^x
Solution:
x^y = y^x
Taking log on both sides
logx^y = logy^x
(i.e.) y log x = x log y
differentiating w.r.to x

Question 5.
(cos x)^{log x}
Solution:
y = (cos x)^{log x}
Taking log on both sides
log y = log (cos x)^{log x} = log x (log cos x)

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Solution:
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
Differentiating w.r.to x

Question 7.
\(\sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)\)
Solution:
\(\sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)\)
Differentiating w.r.to x

Question 8.
tan (x + y) + tan (x – y) = x
Solution:
tan (x + y) + tan (x – y) = x
Differentiating w.r.to x we get

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

Question 10.
\(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:

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

Question 12.
cos [2 \(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)]
Solution:

Question 13.
x = a cos³t; y = a sin²t
Solution:

Question 14.
x = a (cos t + t sin t); y = a [sin t – t cos t]
Solution:

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

Question 16.
\(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:

Question 17.
sin^-1 (3x – 4x³)
Solution:

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

Question 19.
Find the derivative of sin x² with respect to x²
Solution:

Question 20.
Find the derivative of \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) with respect to tan^-1 x.
Solution:
Let u = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) and v = tan^-1 x
Now we have to find \(\frac{d u}{d v}\)

Question 21.

Solution:

Question 22.
Find the derivative with with respect to
Solution:

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

Question 24.
If y = e^tan-1x, show that (1 + x²) y” + (2x – 1) y’ = 0
Solution:
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
Solution:

-xy + (1 – x²) y₁ = 1
differentiating both sides again w.r.to x
-[x y₁ + y (1)] + (1 – x²) (y₂) + y₁ (-2x) = 0
(i.e.) -xy₁ – y + (1 – x²) y₂ – 2xy₁ = 0
(1 – x²) y₂ – 3xy₁ – y = 0

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

Question 27.
If sin y = x sin (a + y) Then prove that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\), a ≠ nπ
Solution:

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

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

Question 1.
If y = A cos4x + B sin 4x, A and B are constants then Show that y₂ + 16y = 0
Solution:

Question 2.
If y = cos (m sin^-1 x), prove that (1 – x²) y₃ – 3xy₂ + (m² – 1) y₁ = 0
Solution:
We have y = cos (m sin^-1 x)
y₁ = sin (m sin^-1x). \(\frac{m}{\sqrt{1-x^{2}}}\)