Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Differentiate the following
Question 1.
y = (x² + 4x + 6)⁵
Solution:
Let = u = x² + 4x + 6
⇒ \(\frac{d u}{d x}\) = 2x + 4
Now y = u⁵ ⇒ \(\frac{d y}{d x}\) = 5u⁴
∴ \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\) = 5u⁴ (2x + 4)
= 5(x² + 4x + 6)⁴ (2x + 4)
= 5 (2x + 4) (x² + 4x + 6)⁴

Question 2.
y = tan 3x
Solution:
y = tan 3x
put u = 3x
\(\frac{d u}{d x}\) = 3
Now y = tan u
⇒ \(\frac{d u}{d x}\) = sec² u
So \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\) = (sec² u) (3)
= 3 sec² 3x

Question 3.
y = cos (tan x)
Solution:
Put u = tan x
\(\frac{d u}{d x}\) = sec²x
Now y = cos u ⇒ \(\frac{d u}{d x}\) = – sin u
Now \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\)
= (-sin u) (sec²x)
= – sec² (sin (tan x))

Question 4.
y = \(\sqrt[3]{1+x^{3}}\)
Solution:

Question 5.
y = \(e^{\sqrt{x}}\)
Solution:

Question 6.
y = sin (e^x)
Solution:
y = sin (e^x)
Let u = e^x

Question 7.
F(x) = (x³ + 4x)⁷
Solution:
F(x) = (x³ + 4x)⁷
Put u = x³ + 4x

Question 8.
h(t) = \(\left(t-\frac{1}{t}\right)^{\frac{3}{2}}\)
Solution:

Question 9.
f(t) = \(\sqrt[3]{1+\tan t}\)
Solution:

Question 10.
y = cos (a³ + x³)
Solution:

Question 11.
y = e^-mx
Solution:

Question 12.
y = 4 sec 5x
Solution:

Question 13.
y = (2x – 5)⁴ (8x² – 5)^-3
Solution:

= \(\frac{8(2 x-5)^{3}}{\left(8 x^{2}-5\right)^{4}}\) (-4x² + 30x – 5)

Question 14.
y = (x² + 1) \(\sqrt[3]{x^{2}+2}\)
Solution:

Question 15.
y = xe^-x2
Solution:
y = xe^-x2
y = uv where u = x and v = e^-x2
Now u’ = 1 and v’ = e^-x2 (-2x)
v’ = – 2xe^-x2
Now y = uv ⇒ y’ = uv’ + vu’
(i.e.) \(\frac{d y}{d x}\) = x[-2xe^-x2] + e^-x2 (1)
= e^-x2 (1 – 2x²)

Question 16.
s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)
Solution:

Question 17.
f(x) = \(\frac{x}{\sqrt{7-3 x}}\)
Solution:

Question 18.
y = tan (cos x)
Solution:
y = tan (cos x)

Question 19.
y = \(\frac{\sin ^{2} x}{\cos x}\)
Solution:

Question 20.
y = \(5^{-\frac{1}{x}}\)
Solution:

Question 21.
y = \(\sqrt{1+2 \tan x}\)
Solution:

Question 22.
y = sin³x + cos³x
Solution:
y = sin³x + cos³x
Here u = sin³ x = (sin x)³
⇒ \(\frac{d u}{d x}\) = 3 (sin x)² (cos x)
= 3sin²x cos x
v = cos³x = (cos x)³
⇒ \(\frac{d v}{d x}\) = 3 (cos x)² (-sin x) = -3 sin x cos²x
Now y = u + v ⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
= 3 sin²x cos x – 3sin x cos²x
= 3 sin x cos x (sin x – cos x)

Question 23.
y = sin² (cos kx)
Solution:

Question 24.
y = (1 + cos²x)⁶
Solution:

Question 25.
y =\(\frac{e^{3 x}}{1+e^{x}}\)
Solution:

Question 26.
y = \(\sqrt{x+\sqrt{x}}\)
Solution:

Question 27.
y = e^{x cos x}
Solution:

Question 28.
y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
Solution:

Question 29.
y = \(\sin (\tan (\sqrt{\sin x}))\)
Solution:

Question 30.
y = sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution: