Class 12 ncert solution math exercise 5.5 differentiation

     EXERCISE 5.5 ( Differentiation )

Differentiate the function with respect to (Class 12 ncert solution math exercise 5.5 differentiation)

Question 1:  x: \cos x \cdot \cos 2 x \cdot \cos 3 x

Solution: Let y=\cos x \cdot \cos 2 x \cdot \cos 3 x

Taking logarithm on both the sides, we obtain

\log y=\log (\cos x \cdot \cos 2 x \cdot \cos 3 x)

\Rightarrow \log y=\log (\cos x)+\log (\cos 2 x)+\log (\cos 3 x)

Differentiating both sides with respect to x, we obtain

\frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\frac{1}{\cos 2 x} \cdot \frac{d}{d x}(\cos 2 x)+\frac{1}{\cos 3 x} \cdot \frac{d}{d x}(\cos 3 x)

\Rightarrow \frac{d y}{d x}=y\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)\right]

\therefore \frac{d y}{d x}=-\cos x \cdot \cos 2 x \cdot \cos 3 x[\tan x+2 \tan 2 x+3 \tan 3 x]

Question 2:Differentiate the function with respect to x: \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}

Solution: Let y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}

Taking logarithm on both the sides, we obtain

\log y=\log \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}

\Rightarrow \log y=\frac{1}{2} \log \left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]

\Rightarrow \log y=\frac{1}{2}[\log \{(x-1)(x-2)\}-\log \{(x-3)(x-4)(x-5)\}]

\Rightarrow \log y=\frac{1}{2}[\log (x-1)+\log (x-2)-\log (x-3)-\log (x-4)-\log (x-5)]

Differentiating both sides with respect to x, we obtain

\frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{l} \frac{1}{x-1} \cdot \frac{d}{d x}(x-1)+\frac{1}{x-2} \cdot \frac{d}{d x}(x-2)-\frac{1}{x-3} \cdot \frac{d}{d x}(x-3)-\frac{1}{x-4} \cdot \frac{d}{d x}(x-4)-\frac{1}{x-5} \cdot \frac{d}{d x}(x-5) \end{array}\right]

\Rightarrow \frac{d y}{d x}=\frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]

\therefore \frac{d y}{d x}=\frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]

Question 3: Differentiate the function with respect to x:(\log x)^{\cos x}

Solution: Let y=(\log x)^{\cos x}

Taking logarithm on both the sides, we obtain

\log y=\cos x \cdot \log (\log x)

Differentiating both sides with respect to x, we obtain

\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\cos x) \cdot \log (\log x)+\cos x \cdot \frac{d}{d x}[\log (\log x)]

\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=-\sin x \log (\log x)+\cos x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)
\Rightarrow \frac{d}{d x}=y\left[-\sin x \log (\log x)+\frac{\cos x}{\log x} \cdot \frac{1}{x}\right]

\therefore \frac{d y}{d x}=(\log x)^{\cos x}\left[\frac{\cos x}{x \log x}-\sin x \log (\log x)\right]

Question 4:Differentiate the function with respect to x: x^x-2^{\sin x}

Solution: Let y=x^x-2^{\sin x}
Also, let x^x=u and 2^{\sin x}=v
\therefore y=u-v

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}

u=x^x

Taking logarithm on both the sides, we obtain

\log u=x \log x

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\left[\frac{d}{d x}(x) \times \log x+x \times \frac{d}{d x}(\log x)\right]

\Rightarrow \frac{d u}{d x}=u\left[1 \times \log x+x \times \frac{1}{x}\right]

\Rightarrow \frac{d u}{d x}=x^x(\log x+1)

\Rightarrow \frac{d u}{d x}=x^x(1+\log x)

v=2^{\sin x}

Taking logarithm on both the sides, we obtain

\log v=\sin x \cdot \log 2

Differentiating both sides with respect to x, we obtain

\frac{1}{v} \cdot \frac{d v}{d x}=\log 2 \cdot \frac{d}{d x}(\sin x)

\Rightarrow \frac{d v}{d x}=v \log 2 \cos x

\Rightarrow \frac{d v}{d x}=2^{\sin x} \cos x \log 2

\therefore \frac{d y}{d x}=x^x(1+\log x)-2^{\sin x} \cos x \log 2

Question 5: Differentiate the function with respect to x:(x+3)^2 \cdot(x+4)^3 \cdot(x+5)^4

Solution: Let y=(x+3)^2 \cdot(x+4)^3 \cdot(x+5)^4

Taking logarithm on both the sides, we obtain

\log y=\log (x+3)^2+\log (x+4)^3+\log (x+5)^4

\Rightarrow \log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5)

Differentiating both sides with respect to x, we obtain

\frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5)

\Rightarrow \frac{d y}{d x}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]

\Rightarrow \frac{d y}{d x}=(x+3)^2(x+4)^3(x+5)^4 \cdot\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]

\Rightarrow \frac{d y}{d x}=(x+3)^2(x+4)^3(x+5)^4\left[\frac{ 2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right]

\Rightarrow \frac{d y}{d x}=(x+3)(x+4)^2(x+5)^3 \cdot\left[2\left(x^2+9 x+20\right)+3\left(x^2+8 x+15\right) \\ +4\left(x^2+7 x+12\right)\right]

\therefore \frac{d y}{d x}=(x+3)(x+4)^2(x+5)^3\left(9 x^2+70 x+133\right)

Question 6: Differentiate the function with respect to x:\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}

Solution: Let y=\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}

Also, let u=\left(x+\frac{1}{x}\right)^x and v=x^{\left(1+\frac{1}{x}\right)}

\therefore y=u+v

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}---(1)

Then, u=\left(x+\frac{1}{x}\right)^x

Taking logarithm on both the sides, we obtain

\Rightarrow \log u=\log \left(x+\frac{1}{x}\right)^x

\Rightarrow \log u=x \log \left(x+\frac{1}{x}\right)

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log \left(x+\frac{1}{x}\right)+x \times \frac{d}{d x}\left[\log \left(x+\frac{1}{x}\right)\right]

\Rightarrow \frac{1}{u} \cdot \frac{d u}{d x}=1 \times \log \left(x+\frac{1}{x}\right)+x \times \frac{1}{\left(x+\frac{1}{x}\right)} \cdot \frac{d}{d x}\left(x+\frac{1}{x}\right)

\Rightarrow \frac{d u}{d x}=u\left(\log \left(x+\frac{1}{x}\right)+\frac{x}{\left(x+\frac{1}{x}\right) \times\left(1-\frac{1}{x^2}\right)}\right]

\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^x\left[\log \left(x+\frac{1}{x}\right)+\frac{\left(x-\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)}\right]

\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^x\left[\log \left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]

\Rightarrow \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right]---(2)

\text { Now, } v=x^{\left(1+\frac{1}{x}\right)}

Taking logarithm on both the sides, we obtain

\Rightarrow \log v=\log \left[x^{\left(1+\frac{1}{x}\right)}\right]

\Rightarrow \log v=\left(1+\frac{1}{x}\right) \log x

Differentiating both sides with respect to x, we obtain

\frac{1}{v} \cdot \frac{d v}{d x}=\left[\frac{d}{d x}\left(1+\frac{1}{x}\right)\right] \times \log x+\left(1+\frac{1}{x}\right) \cdot \frac{d}{d x} \log x

\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=\left(-\frac{1}{x^2}\right) \log x+\left(1+\frac{1}{x}\right) \cdot \frac{1}{x}

\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=-\frac{\log x}{x^2}+\frac{1}{x}+\frac{1}{x^2}

\Rightarrow \frac{d v}{d x}=v\left[\frac{-\log x+x+1}{x^2}\right]

\Rightarrow \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{-\log x+x+1}{x^2}\right]--(3)

Therefore, from (1), (2) and (3);

\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^2}\right]

Question 7: Differentiate the function with respect to x:(\log x)^x+x^{\log x}

Solution: Let y=(\log x)^x+x^{\log x}

\therefore y=u+v

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}

Then, u=(\log x)^x

Taking logarithm on both the sides, we obtain

\Rightarrow \log u=\log \left[(\log x)^x\right]

\Rightarrow \log u=x \log (\log x)

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\log x)+x \cdot \frac{d}{d x}[\log (\log x)]

\Rightarrow \frac{d u}{d x}=u\left[1 \times \log (\log x)+x \cdot \frac{1}{(\log x)} \cdot \frac{d}{d x}(\log x)\right]

\Rightarrow \frac{d u}{d x}=(\log x)^x\left[\log (\log x)+\frac{x}{(\log x)} \cdot \frac{1}{x}\right]

\Rightarrow \frac{d u}{d x}=(\log x)^x\left[\log (\log x)+\frac{1}{(\log x)}\right]

\Rightarrow \frac{d u}{d x}=(\log x)^x\left[\frac{\log (\log x) \cdot \log x+1}{\log x}\right]

\Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]

v=x^{\log x}

Taking logarithm on both the sides, we obtain

\Rightarrow \log v=\log \left(x^{\log x}\right)

\Rightarrow \log v=\log x \log x=(\log x)^2

Differentiating both sides with respect to x, we obtain

\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}\left[(\log x)^2\right]

\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=2(\log x) \cdot \frac{d}{d x}(\log x)

\Rightarrow \frac{d v}{d x}=2 v(\log x) \cdot \frac{1}{x}

\Rightarrow \frac{d v}{d x}=2 x^{\log x} \frac{\log x}{x}

&\Rightarrow \frac{d v}{d x}=2 x^{\log x-1} \cdot \log x

Therefore, from (1), (2) and (3);

\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 x^{\log x-1} \cdot \log x

Question 8:Differentiate the function with respect to x:(\sin x)^x+\sin ^{-1} \sqrt{x}

Solution: Let y=(\sin x)^x+\sin ^{-1} \sqrt{x}

Also, let u=(\sin x)^x and v=\sin ^{-1} \sqrt{x}

\therefore y=u+v ---(1)

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}

Then, u=(\sin x)^x

Taking logarithm on both the sides, we obtain

\Rightarrow \log u=\log (\sin x)^x

\Rightarrow \log u=x \log (\sin x)

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\sin x)+x \cdot \frac{d}{d x}[\log (\sin x)]

\Rightarrow \frac{d u}{d x}=u\left[1 \times \log (\sin x)+x \cdot \frac{1}{(\sin x)} \cdot \frac{d}{d x}(\sin x)\right]

\Rightarrow \frac{d u}{d x}=(\sin x)^x\left[\log (\sin x)+\frac{x}{(\sin x)} \cdot \cos x\right]

\Rightarrow \frac{d u}{d x}=(\sin x)^x[x \cot x+\log \sin x]---(2)

v=\sin ^{-1} \sqrt{x}

Differentiating both sides with respect to x, we obtain

\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{d x}(\sqrt{x})

\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}

\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^2}}---(3)

Therefore, from (1), (2) and (3);

\frac{d y}{d x}=(\sin x)^x[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^2}}

Question 9:Differentiate the function with respect to x: x^{\sin x}+(\sin x)^{\cos x}

Solution: Let y=x^{\sin x}+(\sin x)^{\cos x}

Also, let u=x^{\sin x} and v=(\sin x)^{\cos x}

\therefore y=u+v

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} --(1)

Then, u=x^{\sin x}

Taking logarithm on both the sides, we obtain

\Rightarrow \log u=\log \left(x^{\sin x}\right)

\Rightarrow \log u=\sin x \log x

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(\sin x) \cdot \log x+\sin x \cdot \frac{d}{d x}(\log x)

\Rightarrow \frac{d u}{d x}=u\left[\cos x \log x+\sin x \cdot \frac{1}{x}\right]

\Rightarrow \frac{d u}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]---(2)

v=(\sin x)^{\cos x}

Taking logarithm on both the sides, we obtain

\Rightarrow \log v=\log (\sin x)^{\cos x}

\Rightarrow \log v=\cos x \log (\sin x)

Differentiating both sides with respect to x, we obtain

\frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}(\cos x) \times \log (\sin x)+\cos x \times \frac{d}{d x}[\log (\sin x)]

\Rightarrow \frac{d v}{d x}=v\left[-\sin x \cdot \log (\sin x)+\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right]

\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}\left[-\sin x \log (\sin x)+\frac{\cos x}{\sin x} \cos x\right]

\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[-\sin x \log (\sin x)+\cot x \cos x]---(3)

\Rightarrow \frac{d v}{d x}=(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)]

Therefore, from (1),(2) and (3)

\frac{d y}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]+(\sin x)^{\cos x}[\cot x \cos x-\sin x \log (\sin x)]

Question 10:Differentiate the function with respect to x: x^{x^{x \cos x}+\frac{x^2+1}{x^2-1}}

Solution:Let y=x^{x \cos x}+\frac{x^2+1}{x^2-1}

Also, let u=x^{x \cos x} and v=\frac{x^2+1}{x^2-1}

\therefore y=u+v

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}

Then, u=x^{x \cos x}

Taking logarithm on both the sides, we obtain

\Rightarrow \log u=\log \left(x^{x \cos x}\right)

\Rightarrow \log u=x \cos x \log x

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x) \cdot \cos x \cdot \log x+x \cdot \frac{d}{d x}(\cos x) \cdot \log x+x \cos x \cdot \frac{d}{d x}(\log x)

\Rightarrow \frac{d u}{d x}=u\left[1 \cdot \cos x \cdot \log x+x \cdot(-\sin x) \log x+x \cos x \cdot \frac{1}{x}\right]

\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x \log x-x \cdot \sin x \log x+\cos x]

\Rightarrow \frac{d u}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \sin x \log x]---(2)

v=\frac{x^2+1}{x^2-1}

Taking logarithm on both the sides, we obtain

\Rightarrow \log v=\log \left(x^2+1\right)-\log \left(x^2-1\right)

Differentiating both sides with respect to x, we obtain

\frac{1}{v} \frac{d v}{d x}=\frac{2 x}{x^2+1}-\frac{2 x}{x^2-1}

\Rightarrow \frac{d v}{d x}=v\left[\frac{2 x\left(x^2-1\right)-2 x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}\right]

&\Rightarrow \frac{d v}{d x}=\frac{x^2+1}{x^2-1} \times\left[\frac{-4 x}{\left(x^2+1\right)\left(x^2-1\right)}\right]

\Rightarrow \frac{d v}{d x}=\frac{-4 x}{\left(x^2-1\right)^2}---(3)

Therefore, from (1),(2) and (3);

\frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \sin x \log x]-\frac{4 x}{\left(x^2-1\right)^2}

Question 11: Differentiate the function with respect to x:(x \cos x)^x+(x \sin x)^{\frac{1}{x}}

Solution: Let y=(x \cos x)^x+(x \sin x)^{\frac{1}{x}}
Also, let u=(x \cos x)^x and v=(x \sin x)^{\frac{1}{x}}

\therefore y=u+v

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}---(1)

Then, u=(x \cos x)^x

Taking logarithm on both the sides, we obtain

\Rightarrow \log u=(x \cos x)^x

\Rightarrow \log u=x \log (x \cos x)

\Rightarrow \log u=x[\log x+\log \cos x]

\Rightarrow \log u=x \log x+x \log \cos x

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x \log x)+\frac{d}{d x}(x \log \cos x)

\Rightarrow \frac{d u}{d x}=u\left[\left\{\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)\right\}+\left\{\log \cos x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos x)\right\}\right]

\Rightarrow \frac{d u}{d x}=(x \cos x)^x\left[\left(\log x \cdot 1+x \cdot \frac{1}{x}\right)+\left\{\log \cos x \cdot 1+x \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)\right\}\right]

\Rightarrow \frac{d u}{d x}=(x \cos x)^x\left[(\log x+1)+\left\{\log \cos x+\frac{x}{\cos x} \cdot(-\sin x)\right\}\right]

\Rightarrow \frac{d u}{d x}=(x \cos x)^x[(1+\log x)+(\log \cos x-x \tan x)]

\Rightarrow \frac{d u}{d x}=(x \cos x)^x[(1-x \tan x)+(\log x+\log \cos x)]

\Rightarrow \frac{d u}{d x}=(x \cos x)^x[1-x \tan x+\log (x \cos x)]---(2)

v=(x \sin x)^{\frac{1}{x}}

Taking logarithm on both the sides, we obtain

\Rightarrow \log v=\log (x \sin x)^{\frac{1}{x}}

\Rightarrow \log v=\frac{1}{x} \log (x \sin x)

\Rightarrow \log v=\frac{1}{x}(\log x+\log \sin x)

\Rightarrow \log v=\frac{1}{x} \log x+\frac{1}{x} \log \sin x

Differentiating both sides with respect to x, we obtain

\frac{1}{v} \frac{d v}{d x}=\frac{d}{d x}\left(\frac{1}{x} \log x\right)+\frac{d}{d x}\left[\frac{1}{x} \log (\sin x)\right]

\Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{1}{x} \cdot \frac{d}{d x}(\log x)\right]+\left[\log (\sin x) \cdot \frac{d}{d x}\left(\frac{1}{x}\right)+\frac{1}{x} \cdot \frac{d}{d x}\{\log (\sin x)\}\right]

\Rightarrow \frac{1}{v} \frac{d v}{d x}=\left[\log x \cdot\left(-\frac{1}{x^2}\right)+\frac{1}{x} \cdot \frac{1}{x}\right]+\left[\log (\sin x) \cdot\left(-\frac{1}{x^2}\right)+\frac{1}{x} \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right]

\Rightarrow \frac{1}{v} \frac{d v}{d x}=\frac{1}{x^2}(1-\log x)+\left[-\frac{\log (\sin x)}{x^2}+\frac{1}{x \sin x} \cdot \cos x\right]

\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x}{x^2}+\frac{-\log (\sin x)+x \cot x}{x^2}\right]

\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x-\log (\sin x)+x \cot x}{x^2}\right]

\Rightarrow \frac{d v}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log (x \sin x)+x \cot x}{x^2}\right]---(3)

Therefore, from (1), (2) and (3);

\frac{d y}{d x}=(x \cos x)^x[1-x \tan x+\log (x \cos x)]+(x \sin x)^{\frac{1}{x}}\left[\frac{x \cot x+1-\log (x \sin x)}{x^2}\right]

Question 12: Find \frac{d y}{d x} of the function x^y+y^x=1

Solution:The given function is x^y+y^x=1

Let, x^y=u and y^x=v

\therefore u+v=1

\Rightarrow \frac{d u}{d x}+\frac{d v}{d x}=0 ---(1)

Then, u=x^y

Taking logarithm on both the sides, we obtain

\Rightarrow \log u=\log \left(x^y\right)

\Rightarrow \log u=y \log x

Differentiating both sides with respect to x, we obtain

\frac{1}{u} \cdot \frac{d u}{d x}=\log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x)

\Rightarrow \frac{d u}{d x}=u\left[\log x \frac{d y}{d x}+y \cdot \frac{1}{x}\right]

\Rightarrow \frac{d u}{d x}=x^y\left[\log x \frac{d y}{d x}+\frac{y}{x}\right]---(2)

Now, v=y^x

Taking logarithm on both the sides, we obtain

\Rightarrow \log v=\log \left(y^x\right)

\Rightarrow \log v=x \log y

Differentiating both sides with respect to x, we obtain

\frac{1}{v} \frac{d v}{d x}=\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)

\Rightarrow \frac{d v}{d x}=v\left[\log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right]

\Rightarrow \frac{d v}{d x}=y^x\left[\log y+\frac{x}{y} \cdot \frac{d y}{d x}\right]---(3)

Therefore, from (1), (2) and (3);

x^y\left[\log x \frac{d y}{d x}+\frac{y}{x}\right]+y^x\left[\log y+\frac{x}{y} \cdot \frac{d y}{d x}\right]=0

\Rightarrow\left(x^y \log x+x y^{x-1}\right) \frac{d y}{d x}=-\left(y x^{y-1}+y^x \log y\right)

\therefore \frac{d y}{d x}=\frac{-\left(y x^{y-1}+y^x \log y\right)}{\left(x^y \log x+x y^{x-1}\right)}

Question 13: Find \frac{d y}{d x} of the function y^x=x^y

Solution: The given function is y^x=x^y

Taking logarithm on both the sides, we obtain

x \log y=y \log x

Differentiating both sides with respect to x, we obtain

\log y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log y)=\log x \cdot \frac{d}{d x}(y)+y \cdot \frac{d}{d x}(\log x)

\Rightarrow \log y \cdot 1+x \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+y \cdot \frac{1}{x}

\Rightarrow \log y+\frac{x}{y} \cdot \frac{d y}{d x}=\log x \cdot \frac{d y}{d x}+\frac{y}{x}

\Rightarrow\left(\frac{x}{y}-\log x\right) \frac{d y}{d x}=\frac{y}{x}-\log y

\Rightarrow\left(\frac{x-y \log x}{y}\right) \frac{d y}{d x}=\frac{y-x \log y}{x}

\therefore \frac{d y}{d x}=\frac{y}{x}\left(\frac{y-x \log y}{x-y \log x}\right)

Question 14: Find \frac{d y}{d x} of the function (\cos x)^y=(\cos y)^x

Solution: The given function is (\cos x)^y=(\cos y)^x

Taking logarithm on both the sides, we obtain

y \log \cos x=x \log \cos y

Differentiating both sides with respect to x, we obtain

\log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log \cos x)=\log \cos y \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log \cos y)

\Rightarrow \log \cos x \cdot \frac{d y}{d x}+y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)=\log \cos y \cdot 1+x \cdot \frac{1}{\cos y} \cdot \frac{d}{d x}(\cos y)

\Rightarrow \log \cos x \cdot \frac{d y}{d x}+\frac{y}{\cos x} \cdot(-\sin x)=\log \cos y+\frac{x}{\cos y} \cdot(-\sin y) \cdot \frac{d y}{d x}

\Rightarrow \log \cos x \cdot \frac{d y}{d x}-y \tan x=\log \cos y-x \tan y \frac{d y}{d x}

\Rightarrow(\log \cos x+x \tan y) \frac{d y}{d x}=y \tan x+\log \cos y

\therefore \frac{d y}{d x}=\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}

Question 15: Find \frac{d y}{d x} of the function x y=e^{(x-y)}

Solution: The given function is x y=e^{(x-y)}

Taking logarithm on both the sides, we obtain

\log (x y)=\log \left(e^{x-y}\right)

\Rightarrow \log x+\log y=(x-y) \log e

\Rightarrow \log x+\log y=(x-y) \times 1

\Rightarrow \log x+\log y=(x-y)

Differentiating both sides with respect to x, we obtain

\frac{d}{d x}(\log x)+\frac{d}{d x}(\log y)=\frac{d}{d x}(x)-\frac{d y}{d x}

\Rightarrow \frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}

\Rightarrow\left(1+\frac{1}{y}\right) \frac{d y}{d x}=1-\frac{1}{x}

\Rightarrow\left(\frac{y+1}{y}\right) \frac{d y}{d x}=\frac{x-1}{x}

\therefore \frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}

Question 16: Find the derivative of the function given by f(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right) and hence find f^{\prime}(1)

Solution: The given function is f(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)

Taking logarithm on both the sides, we obtain

\log f(x)=\log (1+x)+\log \left(1+x^2\right)+\log \left(1+x^4\right)+\log \left(1+x^8\right)

Differentiating both sides with respect to x, we obtain

\frac{1}{f(x)} \cdot \frac{d}{d x}[f(x)]=\frac{d}{d x} \log (1+x)+\frac{d}{d x} \log \left(1+x^2\right)

+\frac{d}{d x} \log \left(1+x^4\right)+\frac{d}{d x} \log \left(1+x^8\right)

\Rightarrow \frac{1}{f(x)} \cdot f^{\prime}(x)= \frac{1}{1+x} \cdot \frac{d}{d x}(1+x)+\frac{1}{1+x^2} \cdot \frac{d}{d x}\left(1+x^2\right)

+\frac{1}{1+x^4} \cdot \frac{d}{d x}\left(1+x^4\right)+\frac{1}{1+x^8} \cdot \frac{d}{d x}\left(1+x^8\right)

\Rightarrow f^{\prime}(x)=f(x)\left[\frac{1}{1+x}+\frac{1}{1+x^2} \cdot 2 x+\frac{1}{1+x^4} \cdot 4 x^3+\frac{1}{1+x^8} \cdot 8 x^7\right]

\therefore f^{\prime}(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left[\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}\right]

Hence,

f^{\prime}(1) =(1+1)\left(1+1^2\right)\left(1+1^4\right)\left(1+1^8\right)\left[\frac{1}{1+1}+\frac{2(1)}{1+1^2}+\frac{4(1)^3}{1+1^4}+\frac{8(1)^7}{1+1^8}\right]

=2 \times 2 \times 2 \times 2\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]

=16\left(\frac{15}{2}\right)

=120

Question 17:Differentiate\left(x^2-5 x+8\right)\left(x^3+7 x+9\right) in three ways mentioned below.
(i) By using product rule
(ii) By expanding the product to obtain a single polynomial.
(iii) By logarithmic differentiation.
Do they all give the same answer?

Solution: Let y=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)

(i) By using product rule

Let u=\left(x^2-5 x+8\right) and v=x^3+7 x+9

\therefore y=u v

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x}

\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^2-5 x+8\right) \cdot\left(x^3+7 x+9\right)+\left(x^2-5 x+8\right) \cdot \frac{d}{d x}\left(x^3+7 x+9\right)

\Rightarrow \frac{d y}{d x}=(2 x-5)\left(x^3+7 x+9\right)+\left(x^2-5 x+8\right)\left(3 x^2+7\right)

\Rightarrow \frac{d y}{d x}=2 x\left(x^3+7 x+9\right)-5\left(x^3+7 x+9\right)+x^2\left(3 x^2+7\right)

-5 x\left(3 x^2+7\right)+8\left(3 x^2+7\right)

\Rightarrow \frac{d y}{d x}=\left(2 x^4+14 x^2+18 x\right)-5 x^3-35 x-45+\left(3 x^4+7 x^2\right)

\Rightarrow \frac{d y}{d x}=-15 x^3-35 x+24 x^2+56

\therefore \frac{d y}{d x}=5 x^4-20 x^3+45 x^2-52 x+11

(ii) By expanding the product to obtain a single polynomial.

y=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)

=x^2\left(x^3+7 x+9\right)-5 x\left(x^3+7 x+9\right)+8\left(x^3+7 x+9\right)

=x^5+7 x^3+9 x^2-5 x^4-35 x^2-45 x+8 x^3+56 x+72

=x^5-5 x^4+15 x^3-26 x^2+11 x+72

Therefore,

\frac{d y}{d x}=\frac{d}{d x}\left(x^5-5 x^4+15 x^3-26 x^2+11 x+72\right)

=\frac{d}{d x}\left(x^5\right)-5 \frac{d}{d x}\left(x^4\right)+15 \frac{d}{d x}\left(x^3\right)-26 \frac{d}{d x}\left(x^2\right)+11 \frac{d}{d x}(x)+\frac{d}{d x}(72)

=5 x^4-5\left(4 x^3\right)+15\left(3 x^2\right)-26(2 x)+11(1)+0

=5 x^4-20 x^3+45 x^2-52 x+11

(iii) By logarithmic differentiation.

y=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)

Taking logarithm on both the sides, we obtain

\log y=\log \left(x^2-5 x+8\right)+\log \left(x^3+7 x+9\right)

Differentiating both sides with respect to x, we obtain

\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x} \log \left(x^2-5 x+8\right)+\frac{d}{d x} \log \left(x^3+7 x+9\right)

\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{x^2-5 x+8} \cdot \frac{d}{d x}\left(x^2-5 x+8\right)+\frac{1}{x^3+7 x+9} \cdot \frac{d}{d x}\left(x^3+7 x+9\right)

\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x^2-5 x+8} \cdot(2 x-5)+\frac{1}{x^3+7 x+9} \cdot\left(3 x^2+7\right)\right]

\Rightarrow \frac{d y}{d x}=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)\left[\frac{2 x-5}{x^2-5 x+8}+\frac{3 x^2+7}{x^3+7 x+9}\right]

\Rightarrow \frac{d y}{d x}=\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)\left[\frac{(2 x-5)\left(x^3+7 x+9\right)+\left(3 x^2+7\right)\left(x^2-5 x+8\right)}{\left(x^2-5 x+8\right)\left(x^3+7 x+9\right)}\right]

\Rightarrow \frac{d y}{d x}=2 x\left(x^3+7 x+9\right)-5\left(x^3+7 x+9\right)+3 x^2\left(x^2-5 x+8\right)+7\left(x^2-5 x+8\right)

\Rightarrow \frac{d y}{d x}=2 x^4+14 x^2+18 x-5 x^3-35 x-45+3 x^5-15 x^3+24 x^2+7 x^2-35 x+56

\Rightarrow \frac{d y}{d x}=5 x^4-20 x^3+45 x^2-52 x+11

From the above three observations, it can be concluded that

all the results of \frac{d y}{d x} are same.

Question 18:If u, v and w are functions of x, then show that

\frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}

in two ways – first by repeated application of product rule, second by logarithmic differentiation.

Solution: Let y=u . v . w=u .(v . w)

By applying product rule, we get

\frac{d y}{d x}=\frac{d u}{d x} \cdot(v \cdot w)+u \cdot \frac{d}{d x}(v \cdot w)

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot(v \cdot w)+u\left[\frac{d v}{d x} \cdot w+v \cdot \frac{d w}{d x}\right] \quad \text { (Again applying product rule) }

\Rightarrow \frac{d y}{d x}=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}

Taking logarithm on both the sides of the equation y=u . v . w,
we obtain, \log y=\log u+\log v+\log w

Differentiating both sides with respect to x, we obtain
\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(\log u)+\frac{d}{d x}(\log v)+\frac{d}{d x}(\log w)

\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}

\Rightarrow \frac{d y}{d x}=y\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right)

\Rightarrow \frac{d y}{d x}=u \cdot v \cdot w\left(\frac{1}{u} \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \frac{d w}{d x}\right)

\therefore \frac{d}{d x}(u . v . w)=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x}

 

 

 

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