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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 cos^{3}t; y = a sin^{2}t

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 – 4x^{3})

Solution:

Question 18.

\(\tan ^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\)

Solution:

Question 19.

Find the derivative of sin x^{2} with respect to x^{2}

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 + x^{2}) y” + (2x – 1) y’ = 0

Solution:

y = e^{tan-1}x

y’ = e^{tan-1} × \(\frac{1}{1+x^{2}}\)

(1 + x^{2}) y’ = e^{tan-1}x

(1 + x^{2}) y’ = y

Differentiating with respect to x

(1 + x^{2}) y” + y’ (0 + 2x) = y’

(1 + x^{2}) y” + 2xy’ – y’ = 0

(1 + x^{2}) y” + (2x – 1) y’ = 0

Question 25.

If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) show that (1 – x^{2}) y_{2} – 3xy_{1} – y = 0

Solution:

-xy + (1 – x^{2}) y_{1} = 1

differentiating both sides again w.r.to x

-[x y_{1} + y (1)] + (1 – x^{2}) (y_{2}) + y_{1} (-2x) = 0

(i.e.) -xy_{1} – y + (1 – x^{2}) y_{2} – 2xy_{1} = 0

(1 – x^{2}) y_{2} – 3xy_{1} – 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 – x^{2}) \(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}\) – 2 = 0. Hence find y_{2} 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 cos 4x + B sin 4x, A and B are constants then Show that y_{2} + 16y = 0

Solution:

Question 2.

If y = cos (m sin^{-1} x), prove that (1 – x^{2}) y_{3} – 3xy_{2} + (m^{2} – 1) y_{1} = 0

Solution:

We have y = cos (m sin^{-1} x)

y_{1} = sin (m sin^{-1}x). \(\frac{m}{\sqrt{1-x^{2}}}\)