Exercise 5.5

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Q.1 Differentiate

$$ y=\cos x\cdot\cos2x\cdot\cos3x $$

with respect to $x$.

Solution:

Given,

$$ y=\cos x\cdot\cos2x\cdot\cos3x. $$

Since the function is the product of three functions, we use logarithmic differentiation.

Taking logarithm on both sides,

$$ \log y = \log(\cos x) + \log(\cos2x) + \log(\cos3x). $$

Differentiating both sides with respect to $x$,

$$ \begin{aligned} \frac1y\frac{dy}{dx} &= -\tan x - 2\tan2x - 3\tan3x. \end{aligned} $$

Multiplying both sides by $y$,

$$ \begin{aligned} \frac{dy}{dx} &= \cos x\cos2x\cos3x \left( -\tan x - 2\tan2x - 3\tan3x \right). \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = -\cos x\cos2x\cos3x \left( \tan x + 2\tan2x + 3\tan3x \right) } $$

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Q.2 Differentiate

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

with respect to $x$.

Solution:

Given,

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

Using logarithmic differentiation,

$$ \begin{aligned} \log y &= \frac12 \left[ \log(x-1) +\log(x-2) -\log(x-3) -\log(x-4) -\log(x-5) \right]. \end{aligned} $$

Differentiating both sides,

$$ \begin{aligned} \frac1y\frac{dy}{dx} &= \frac12 \left( \frac1{x-1} +\frac1{x-2} -\frac1{x-3} -\frac1{x-4} -\frac1{x-5} \right). \end{aligned} $$

Multiplying both sides by $y$,

$$ \boxed{ \frac{dy}{dx} = \frac12 \sqrt{ \frac{(x-1)(x-2)} {(x-3)(x-4)(x-5)} } \left( \frac1{x-1} +\frac1{x-2} -\frac1{x-3} -\frac1{x-4} -\frac1{x-5} \right) } $$

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Q.3 Differentiate

$$ y=(\log x)^{\cos x} $$

with respect to $x$.

Solution:

Given,

$$ y=(\log x)^{\cos x}. $$

Since both the base and the exponent are functions of $x$, we use logarithmic differentiation.

Taking logarithm on both sides,

$$ \log y = \cos x\, \log(\log x). $$

Differentiating both sides,

$$ \begin{aligned} \frac1y\frac{dy}{dx} &= -\sin x\,\log(\log x) + \cos x\cdot \frac1{\log x} \cdot \frac1x. \end{aligned} $$

Hence,

$$ \begin{aligned} \frac{dy}{dx} &= (\log x)^{\cos x} \left[ -\sin x\,\log(\log x) + \frac{\cos x}{x\log x} \right]. \end{aligned} $$

Answer:

$$ \boxed{ \frac{dy}{dx} = (\log x)^{\cos x} \left[ -\sin x\,\log(\log x) + \frac{\cos x}{x\log x} \right] } $$

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Q.4 Differentiate

$$ y=x^x-2^{\sin x} $$

with respect to $x$.

Solution:

Given,

$$ y=x^x-2^{\sin x}. $$

Differentiate each term separately.

First term: $x^x$

Using logarithmic differentiation,

$$ \begin{aligned} u&=x^x,\\ \log u&=x\log x. \end{aligned} $$

Differentiating,

$$ \begin{aligned} \frac1u\frac{du}{dx} &=\log x+1. \end{aligned} $$

Hence,

$$ \frac{du}{dx} = x^x(\log x+1). $$

Second term: $2^{\sin x}$

Using

$$ \frac{d}{dx}(a^u)=a^u\log a\cdot\frac{du}{dx}, $$

where $u=\sin x$,

$$ \frac{du}{dx}=\cos x. $$

Therefore,

$$ \frac{d}{dx}(2^{\sin x}) = 2^{\sin x}\log2\cos x. $$

Hence,

$$ \boxed{ \frac{dy}{dx} = x^x(\log x+1) - 2^{\sin x}\log2\cos x } $$

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Q.5 Differentiate

$$ y=(x+3)^2(x+4)^3(x+5)^4 $$

with respect to $x$.

Solution:

Given,

$$ y=(x+3)^2(x+4)^3(x+5)^4. $$

Using logarithmic differentiation,

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

Differentiating,

$$ \begin{aligned} \frac1y\frac{dy}{dx} &= \frac2{x+3} + \frac3{x+4} + \frac4{x+5}. \end{aligned} $$

Multiplying by $y$,

$$ \boxed{ \frac{dy}{dx} = (x+3)^2(x+4)^3(x+5)^4 \left( \frac2{x+3} + \frac3{x+4} + \frac4{x+5} \right) } $$

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Q.6 Differentiate

$$ y= \left(x+\frac1x\right)^x + x^{\left(1+\frac1x\right)} $$

with respect to $x$.

Solution:

Differentiate each term separately.

First term:

$$ u=\left(x+\frac1x\right)^x. $$

Taking logarithm,

$$ \log u = x\log\left(x+\frac1x\right). $$

Differentiating,

$$ \begin{aligned} \frac1u\frac{du}{dx} &= \log\left(x+\frac1x\right) + x \cdot \frac{1-\frac1{x^2}} {x+\frac1x}. \end{aligned} $$

Since

$$ x+\frac1x=\frac{x^2+1}{x}, $$

we obtain

$$ \begin{aligned} x \cdot \frac{1-\frac1{x^2}} {x+\frac1x} &= \frac{x^2-1}{x^2+1}. \end{aligned} $$

Hence,

$$ \frac{du}{dx} = \left(x+\frac1x\right)^x \left[ \log\left(x+\frac1x\right) + \frac{x^2-1}{x^2+1} \right]. $$

Second term:

$$ v=x^{\left(1+\frac1x\right)}. $$

Taking logarithm,

$$ \log v = \left(1+\frac1x\right)\log x. $$

Differentiating,

$$ \begin{aligned} \frac1v\frac{dv}{dx} &= -\frac{\log x}{x^2} + \left(1+\frac1x\right)\frac1x\\[2mm] &= \frac1x+\frac1{x^2}-\frac{\log x}{x^2}. \end{aligned} $$

Therefore,

$$ \frac{dv}{dx} = x^{\left(1+\frac1x\right)} \left( \frac1x+\frac1{x^2}-\frac{\log x}{x^2} \right). $$

Finally,

$$ \boxed{ \begin{aligned} \frac{dy}{dx} &= \left(x+\frac1x\right)^x \left[ \log\left(x+\frac1x\right) + \frac{x^2-1}{x^2+1} \right] \\[2mm] &\quad+ x^{\left(1+\frac1x\right)} \left( \frac1x+\frac1{x^2}-\frac{\log x}{x^2} \right). \end{aligned} } $$

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Q.7 Differentiate

$$ y=(\log x)^x+x^{\log x} $$

with respect to $x$.

Solution:

Given,

$$ y=(\log x)^x+x^{\log x}. $$

Differentiate each term separately using logarithmic differentiation.

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First term:

$$ u=(\log x)^x. $$

Taking logarithm on both sides,

$$ \log u=x\log(\log x). $$

Differentiating,

$$ \begin{aligned} \frac1u\frac{du}{dx} &=\log(\log x) +x\cdot\frac1{\log x}\cdot\frac1x\\ &=\log(\log x)+\frac1{\log x}. \end{aligned} $$

Hence,

$$ \frac{du}{dx} = (\log x)^x \left( \log(\log x)+\frac1{\log x} \right). $$

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Second term:

$$ v=x^{\log x}. $$

Taking logarithm,

$$ \log v=(\log x)^2. $$

Differentiating,

$$ \begin{aligned} \frac1v\frac{dv}{dx} &= 2(\log x)\cdot\frac1x. \end{aligned} $$

Therefore,

$$ \frac{dv}{dx} = x^{\log x} \cdot \frac{2\log x}{x}. $$

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Hence,

$$ \boxed{ \frac{dy}{dx} = (\log x)^x \left( \log(\log x)+\frac1{\log x} \right) + \frac{2\log x}{x}\, x^{\log x} } $$

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Q.8 Differentiate

$$ y=(\sin x)^x+\sin^{-1}\sqrt{x} $$

with respect to $x$.

Solution:

Given,

$$ y=(\sin x)^x+\sin^{-1}\sqrt{x}. $$

Differentiate each term separately.

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First term:

$$ u=(\sin x)^x. $$

Taking logarithm,

$$ \log u=x\log(\sin x). $$

Differentiating,

$$ \begin{aligned} \frac1u\frac{du}{dx} &= \log(\sin x) + x\cdot\frac{\cos x}{\sin x}\\ &= \log(\sin x)+x\cot x. \end{aligned} $$

Hence,

$$ \frac{du}{dx} = (\sin x)^x \left( \log(\sin x)+x\cot x \right). $$

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Second term:

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

Using the chain rule,

$$ \frac{dv}{dx} = \frac1{\sqrt{1-(\sqrt{x})^2}} \cdot \frac1{2\sqrt{x}}. $$

Since

$$ (\sqrt{x})^2=x, $$

we obtain

$$ \frac{dv}{dx} = \frac1{2\sqrt{x}\sqrt{1-x}}. $$

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Therefore,

$$ \boxed{ \frac{dy}{dx} = (\sin x)^x \left( \log(\sin x)+x\cot x \right) + \frac1{2\sqrt{x}\sqrt{1-x}} } $$

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Q.9 Differentiate

$$ y=x^{\sin x}+(\sin x)^{\cos x} $$

with respect to $x$.

Solution:

Differentiate each term separately.

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First term:

$$ u=x^{\sin x}. $$

Taking logarithm,

$$ \log u=\sin x\log x. $$

Differentiating,

$$ \begin{aligned} \frac1u\frac{du}{dx} &= \cos x\log x +\frac{\sin x}{x}. \end{aligned} $$

Hence,

$$ \frac{du}{dx} = x^{\sin x} \left( \cos x\log x +\frac{\sin x}{x} \right). $$

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Second term:

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

Taking logarithm,

$$ \log v=\cos x\log(\sin x). $$

Differentiating,

$$ \begin{aligned} \frac1v\frac{dv}{dx} &= -\sin x\log(\sin x) + \cos x\cdot\cot x. \end{aligned} $$

Therefore,

$$ \frac{dv}{dx} = (\sin x)^{\cos x} \left( -\sin x\log(\sin x) +\cos x\cot x \right). $$

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Hence,

$$ \boxed{ \begin{aligned} \frac{dy}{dx} &= x^{\sin x} \left( \cos x\log x +\frac{\sin x}{x} \right) \\[2mm] &\quad+ (\sin x)^{\cos x} \left( -\sin x\log(\sin x) +\cos x\cot x \right). \end{aligned} } $$

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Q.10 Differentiate

$$ y=x^{x\cos x}+\frac{x^2+1}{x^2-1} $$

with respect to $x$.

Solution:

Given,

$$ y=x^{x\cos x}+\frac{x^2+1}{x^2-1}. $$

Differentiate each term separately.

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First term:

$$ u=x^{x\cos x}. $$

Using logarithmic differentiation,

$$ \log u=x\cos x\cdot\log x. $$

Differentiating both sides,

$$ \begin{aligned} \frac1u\frac{du}{dx} &= \frac{d}{dx}(x\cos x)\cdot\log x +x\cos x\cdot\frac1x. \end{aligned} $$

Now,

$$ \begin{aligned} \frac{d}{dx}(x\cos x) &= \cos x-x\sin x. \end{aligned} $$

Hence,

$$ \begin{aligned} \frac1u\frac{du}{dx} &= (\cos x-x\sin x)\log x +\cos x. \end{aligned} $$

Therefore,

$$ \boxed{ \frac{du}{dx} = x^{x\cos x} \left[ (\cos x-x\sin x)\log x +\cos x \right] } $$

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Second term:

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

Using the quotient rule,

$$ \frac{dv}{dx} = \frac{(x^2-1)(2x)-(x^2+1)(2x)} {(x^2-1)^2}. $$

Simplifying,

$$ \begin{aligned} \frac{dv}{dx} &= \frac{2x(x^2-1-x^2-1)} {(x^2-1)^2}\\ &= -\frac{4x}{(x^2-1)^2}. \end{aligned} $$

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Hence,

$$ \boxed{ \frac{dy}{dx} = x^{x\cos x} \left[ (\cos x-x\sin x)\log x +\cos x \right] - \frac{4x}{(x^2-1)^2} } $$

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Q.11 Differentiate

$$ y=(x\cos x)^x+(x\sin x)^{\frac1x} $$

with respect to $x$.

Solution:

Differentiate each term separately.

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First term:

$$ u=(x\cos x)^x. $$

Taking logarithm,

$$ \log u = x\log(x\cos x). $$

Differentiating,

$$ \begin{aligned} \frac1u\frac{du}{dx} &= \log(x\cos x) + x\cdot \frac{\cos x-x\sin x} {x\cos x}. \end{aligned} $$

Since

$$ \frac{\cos x-x\sin x} {x\cos x} = \frac1x-\tan x, $$

we obtain

$$ \begin{aligned} \frac1u\frac{du}{dx} &= \log(x\cos x) + 1 - x\tan x. \end{aligned} $$

Hence,

$$ \boxed{ \frac{du}{dx} = (x\cos x)^x \left[ \log(x\cos x) + 1 - x\tan x \right] } $$

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Second term:

$$ v=(x\sin x)^{\frac1x}. $$

Taking logarithm,

$$ \log v = \frac1x\log(x\sin x). $$

Differentiating,

$$ \begin{aligned} \frac1v\frac{dv}{dx} &= -\frac{\log(x\sin x)}{x^2} + \frac1x \left( \frac1x+\cot x \right). \end{aligned} $$

Therefore,

$$ \boxed{ \frac{dv}{dx} = (x\sin x)^{\frac1x} \left[ -\frac{\log(x\sin x)}{x^2} + \frac1{x^2} + \frac{\cot x}{x} \right] } $$

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Hence,

$$ \boxed{ \begin{aligned} \frac{dy}{dx} &= (x\cos x)^x \left[ \log(x\cos x) + 1 - x\tan x \right] \\[2mm] &\quad+ (x\sin x)^{\frac1x} \left[ -\frac{\log(x\sin x)}{x^2} + \frac1{x^2} + \frac{\cot x}{x} \right]. \end{aligned} } $$

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Q.12 Find $\dfrac{dy}{dx}$ if

$$ x^y+y^x=1. $$

Solution:

Given,

$$ x^y+y^x=1. $$

Differentiate both sides with respect to $x$.

Using logarithmic differentiation,

$$ \frac{d}{dx}(u^v) = u^v \left( \frac{dv}{dx}\log u + \frac{v}{u}\frac{du}{dx} \right). $$

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Differentiate $x^y$:

$$ \begin{aligned} \frac{d}{dx}(x^y) &= x^y \left( \frac{dy}{dx}\log x +\frac{y}{x} \right). \end{aligned} $$

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Differentiate $y^x$:

$$ \begin{aligned} \frac{d}{dx}(y^x) &= y^x \left( \log y + \frac{x}{y}\frac{dy}{dx} \right). \end{aligned} $$

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Differentiating the given equation,

$$ x^y \left( \frac{dy}{dx}\log x+\frac{y}{x} \right) + y^x \left( \log y+\frac{x}{y}\frac{dy}{dx} \right) =0. $$

Collecting the terms containing $\dfrac{dy}{dx}$,

$$ \begin{aligned} \left( x^y\log x + \frac{x}{y}y^x \right) \frac{dy}{dx} &= - \left( \frac{yx^y}{x} + y^x\log y \right). \end{aligned} $$

Therefore,

$$ \boxed{ \frac{dy}{dx} = - \frac{ \dfrac{yx^y}{x} + y^x\log y }{ x^y\log x + \dfrac{x}{y}y^x }. } $$

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Q.13 Find $\dfrac{dy}{dx}$ if

$$ y^x=x^y. $$

Solution:

Given,

$$ y^x=x^y. $$

Taking logarithm on both sides,

$$ x\log y = y\log x. $$

Differentiate both sides with respect to $x$.

$$ \begin{aligned} \frac{d}{dx}(x\log y) &= \frac{d}{dx}(y\log x). \end{aligned} $$

Using the product rule,

$$ \begin{aligned} \log y + \frac{x}{y}\frac{dy}{dx} &= \frac{dy}{dx}\log x + \frac{y}{x}. \end{aligned} $$

Collecting the terms containing $\dfrac{dy}{dx}$,

$$ \begin{aligned} \left( \frac{x}{y} - \log x \right) \frac{dy}{dx} &= \frac{y}{x} - \log y. \end{aligned} $$

Hence,

$$ \boxed{ \frac{dy}{dx} = \frac{ \dfrac{y}{x} - \log y }{ \dfrac{x}{y} - \log x }. } $$

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Q.14 Find $\dfrac{dy}{dx}$ if

$$ (\cos x)^y=(\cos y)^x. $$

Solution:

Given,

$$ (\cos x)^y=(\cos y)^x. $$

Taking logarithm on both sides,

$$ y\log(\cos x) = x\log(\cos y). $$

Differentiate both sides with respect to $x$.

$$ \begin{aligned} \frac{dy}{dx}\log(\cos x) -y\tan x &= \log(\cos y) - x\tan y\frac{dy}{dx}. \end{aligned} $$

Collecting the terms containing $\dfrac{dy}{dx}$,

$$ \begin{aligned} \left( \log(\cos x) + x\tan y \right) \frac{dy}{dx} &= \log(\cos y) + y\tan x. \end{aligned} $$

Therefore,

$$ \boxed{ \frac{dy}{dx} = \frac{ \log(\cos y) + y\tan x }{ \log(\cos x) + x\tan y }. } $$

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Q.15 Find $\dfrac{dy}{dx}$ if

$$ xy=e^{(x-y)}. $$

Solution:

Given,

$$ xy=e^{x-y}. $$

Taking natural logarithm on both sides,

$$ \log(xy)=x-y. $$

Using the property of logarithms,

$$ \log x+\log y=x-y. $$

Differentiate both sides with respect to $x$.

$$ \frac1x+\frac1y\frac{dy}{dx} = 1-\frac{dy}{dx}. $$

Collecting the terms containing $\dfrac{dy}{dx}$,

$$ \begin{aligned} \frac1y\frac{dy}{dx} +\frac{dy}{dx} &= 1-\frac1x. \end{aligned} $$

Therefore,

$$ \begin{aligned} \left( \frac{1+y}{y} \right) \frac{dy}{dx} &= \frac{x-1}{x}. \end{aligned} $$

Hence,

$$ \boxed{ \frac{dy}{dx} = \frac{y(x-1)} {x(1+y)}. } $$

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Q.16 Find the derivative of the function

$$ f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8) $$

and hence find $f'(1)$.

Solution:

Given,

$$ f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8). $$

Since the function is the product of several factors, we use logarithmic differentiation.

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Step 1: Take logarithm on both sides.

$$ \log f(x) = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \log(1+x^8). $$

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Step 2: Differentiate both sides with respect to $x$.

$$ \begin{aligned} \frac{f'(x)}{f(x)} &= \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}. \end{aligned} $$

Therefore,

$$ \boxed{ f'(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right). } $$

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Step 3: Find $f(1)$.

$$ \begin{aligned} f(1) &=(1+1)(1+1)(1+1)(1+1)\\ &=2\times2\times2\times2\\ &=16. \end{aligned} $$

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Step 4: Find $\dfrac{f'(1)}{f(1)}$.

$$ \begin{aligned} \frac{f'(1)}{f(1)} &= \frac12 +\frac22 +\frac42 +\frac82\\[2mm] &= \frac12+1+2+4\\[2mm] &= \frac{15}{2}. \end{aligned} $$

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Step 5: Find $f'(1)$.

$$ \begin{aligned} f'(1) &= f(1)\times\frac{15}{2}\\[2mm] &= 16\times\frac{15}{2}\\[2mm] &= 8\times15\\[2mm] &= 120. \end{aligned} $$

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Answer:

The derivative of the given function is

$$ \boxed{ f'(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \left( \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right). } $$

Hence,

$$ \boxed{f'(1)=120.} $$

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Q.17 Differentiate

$$ (x^2-5x+8)(x^3+7x+9) $$

in the following three ways:

1. Using the product rule.
2. By expanding the product to obtain a single polynomial.
3. By logarithmic differentiation.

Do all three methods give the same answer?

Solution:

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(i) Using the Product Rule

Let

$$ u=x^2-5x+8, \qquad v=x^3+7x+9. $$

Then,

$$ \frac{du}{dx}=2x-5, \qquad \frac{dv}{dx}=3x^2+7. $$

By the product rule,

$$ \begin{aligned} \frac{dy}{dx} &= u\frac{dv}{dx} + v\frac{du}{dx}\\[2mm] &= (x^2-5x+8)(3x^2+7) +(x^3+7x+9)(2x-5). \end{aligned} $$

Hence,

$$ \boxed{ \frac{dy}{dx} = (x^2-5x+8)(3x^2+7) +(x^3+7x+9)(2x-5). } $$

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(ii) By Expanding the Product

First expand the expression:

$$ \begin{aligned} y &=(x^2-5x+8)(x^3+7x+9)\\[2mm] &= x^5 -5x^4 +15x^3 -26x^2 +11x +72. \end{aligned} $$

Differentiating term by term,

$$ \begin{aligned} \frac{dy}{dx} &= 5x^4 -20x^3 +45x^2 -52x +11. \end{aligned} $$

Hence,

$$ \boxed{ \frac{dy}{dx} = 5x^4 -20x^3 +45x^2 -52x +11. } $$

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(iii) By Logarithmic Differentiation

Taking logarithm on both sides,

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

Differentiating,

$$ \begin{aligned} \frac1y\frac{dy}{dx} &= \frac{2x-5}{x^2-5x+8} + \frac{3x^2+7}{x^3+7x+9}. \end{aligned} $$

Multiplying both sides by $y$,

$$ \begin{aligned} \frac{dy}{dx} &= (x^2-5x+8)(x^3+7x+9) \left[ \frac{2x-5}{x^2-5x+8} + \frac{3x^2+7}{x^3+7x+9} \right]. \end{aligned} $$

Simplifying,

$$ \begin{aligned} \frac{dy}{dx} &= (x^2-5x+8)(3x^2+7) + (x^3+7x+9)(2x-5). \end{aligned} $$

Expanding this expression gives

$$ \boxed{ \frac{dy}{dx} = 5x^4 -20x^3 +45x^2 -52x +11. } $$

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Conclusion:

All the three methods give the same derivative.

$$ \boxed{ \frac{dy}{dx} = 5x^4 -20x^3 +45x^2 -52x +11. } $$

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Q.18 If $u$, $v$ and $w$ are functions of $x$, then show that

$$ \frac{d}{dx}(uvw) = \frac{du}{dx}\,vw + u\,\frac{dv}{dx}\,w + uv\,\frac{dw}{dx} $$

in two ways:

1. By repeated application of the product rule.
2. By logarithmic differentiation.

Solution:

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Method I: By Repeated Application of the Product Rule

Let

$$ y=uvw. $$

Treat $uv$ as one function and $w$ as another.

Applying the product rule,

$$ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(uv)\cdot w + uv\cdot\frac{dw}{dx}. \end{aligned} $$

Now differentiate $uv$ again using the product rule.

$$ \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}. $$

Substituting this into the previous equation,

$$ \begin{aligned} \frac{dy}{dx} &= \left( u\frac{dv}{dx} + v\frac{du}{dx} \right)w + uv\frac{dw}{dx}. \end{aligned} $$

Expanding,

$$ \boxed{ \frac{dy}{dx} = \frac{du}{dx}\,vw + u\,\frac{dv}{dx}\,w + uv\,\frac{dw}{dx}. } $$

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Method II: By Logarithmic Differentiation

Let

$$ y=uvw. $$

Taking logarithm on both sides,

$$ \log y = \log u + \log v + \log w. $$

Differentiating both sides with respect to $x$,

$$ \begin{aligned} \frac1y\frac{dy}{dx} &= \frac1u\frac{du}{dx} + \frac1v\frac{dv}{dx} + \frac1w\frac{dw}{dx}. \end{aligned} $$

Multiplying both sides by

$$ y=uvw, $$

we obtain

$$ \begin{aligned} \frac{dy}{dx} &= uvw \left( \frac1u\frac{du}{dx} + \frac1v\frac{dv}{dx} + \frac1w\frac{dw}{dx} \right). \end{aligned} $$

Distributing $uvw$,

$$ \boxed{ \frac{dy}{dx} = \frac{du}{dx}\,vw + u\,\frac{dv}{dx}\,w + uv\,\frac{dw}{dx}. } $$

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Conclusion:

Thus, by both repeated application of the product rule and logarithmic differentiation, we obtain the same result:

$$ \boxed{ \frac{d}{dx}(uvw) = \frac{du}{dx}\,vw + u\,\frac{dv}{dx}\,w + uv\,\frac{dw}{dx}. } $$

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