Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(f) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(f)

Find derivatives of the following functions.

Question 1.

x^{x}

Solution:

Let y = x^{x}

Then In y = x . In x

⇒ \(\frac{d}{d x}\)(In y) = \(\frac{d}{d x}\)(x . In x)

⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = In x + x . \(\frac{1}{x}\) = In x + 1 = 1 + In x

⇒ \(\frac{d y}{d x}\) = y (1 + In x) = x^{x} (1 + In x) = x^{x} In (ex)

Question 2.

\(\left(1+\frac{1}{x}\right)^x\)

Solution:

Question 3.

x^{sin x}

Solution:

y = x^{sin x}

⇒ In y = sin x . In x

⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = cos x In x + sin x . \(\frac{1}{x}\)

⇒ \(\frac{d y}{d x}\) = x^{sin x} (cos x . In x + \(\frac{\sin x}{x}\))

Question 4.

(log x)^{tan x}

Solution:

y = (log x)^{tan x}

⇒ log y = tan x . log (log x)

⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = sec^{2} x log (log x) + tan x . \(\frac{1}{\log x}\) . \(\frac{1}{x}\)

⇒ \(\frac{d y}{d x}\) = (log x)^{tan x} {sec^{2} x . log log x + \(\frac{\tan x}{x \log x}\)}

Question 5.

\(2^{\left(2^x\right)}\)

Solution:

y = \(2^{\left(2^x\right)}\)

⇒ In y = 2^{x} . In 2

⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = 2^{x} . In 2 . In 2

⇒ \(\frac{d y}{d x}\) = \(2^{\left(2^x\right)}\) . 2^{x} . (In 2)^{2}

Question 6.

\((1+\sqrt{x})^{x^2}\)

Solution:

Question 7.

\(\left(\sin ^{-1} x\right)^{\sqrt{1-x^2}}\)

Solution:

Question 8.

\((\tan x)^{\log x^3}\)

Solution:

Question 9.

x^{1/x} + (sin x)^{x}

Solution:

Question 10.

(cos x)^{x} + x^{cos x}

Solution:

Question 11.

(x^{2} + 1)^{2/3} (3x + 1)^{1/4} √x

Solution:

Question 12.

\(\frac{(x+1)(x+2)^2(x+3)^3}{(x-1)(x-2)^2(x-3)^3}\)

Solution:

Question 13.

(sin x)^{x} \(\sqrt{\sin x}\left(1+x^2\right)^{\frac{1}{2}+x}\)

Solution:

Question 14.

(sec x + tan x)^{cot x}

Solution:

y = (sec x + tan x)^{cot x}

⇒ In y = cot x . In (sec x + tan x)

⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = -cosec^{2} x . In (sec x + tan x) + cot x . \(\frac{1}{\sec x+\tan x}\) × (sec x . tan x + sec^{2} x)

= -cosec^{2} x . In (sec x + tan x) + cot x . sec x

∴ \(\frac{d y}{d x}\) = y {-cosec^{2} x . In (sec x + tan x) + cosec x}

= (sec x + tan x)^{cot x} {cosec x – cosec^{2} x . In (sec x + tan x)}

Question 15.

(2^{√x})^{1+√x}

Solution: