Complete solutions for Class 12 Maths (NCERT) Exercise 5.5. Master Logarithmic Differentiation to solve complex products, quotients, and functions of the form $f(x)^{g(x)}$. Includes solutions for Implicit Differentiation problems like $x^y + y^x = 1$.
1. $y = \cos x \cdot \cos 2x \cdot \cos 3x$
We use logarithmic differentiation since it’s a product of three terms.
$$\log y = \log (\cos x \cdot \cos 2x \cdot \cos 3x)$$
Using the property $\log(abc) = \log a + \log b + \log c$:
$$\log y = \log(\cos x) + \log(\cos 2x) + \log(\cos 3x)$$
Differentiate both sides w.r.t. $x$:
$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} (-\sin x) + \frac{1}{\cos 2x} (-\sin 2x) \cdot 2 + \frac{1}{\cos 3x} (-\sin 3x) \cdot 3$$
$$\frac{1}{y} \frac{dy}{dx} = -\tan x – 2\tan 2x – 3\tan 3x$$
$$\frac{dy}{dx} = y (-\tan x – 2\tan 2x – 3\tan 3x)$$
$$\mathbf{\frac{dy}{dx} = – \cos x \cos 2x \cos 3x (\tan x + 2\tan 2x + 3\tan 3x)}$$
2. $y = \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}$
Use logarithmic differentiation for the complex rational function:
$$\log y = \log \left( \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right)$$
Using the properties $\log(\frac{a}{b}) = \log a – \log b$ and $\log(ab) = \log a + \log b$:
$$\log y = [\log(x-1) + \log(x-2)] – [\log(x-3) + \log(x-4) + \log(x-5)]$$
Differentiate both sides w.r.t. $x$:
$$\frac{1}{y} \frac{dy}{dx} = \left(\frac{1}{x-1} + \frac{1}{x-2}\right) – \left(\frac{1}{x-3} + \frac{1}{x-4} + \frac{1}{x-5}\right)$$
$$\mathbf{\frac{dy}{dx} = \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \left[ \frac{1}{x-1} + \frac{1}{x-2} – \frac{1}{x-3} – \frac{1}{x-4} – \frac{1}{x-5} \right]}$$
3. $y = (\log x)^{\cos x}$
This is of the form $f(x)^{g(x)}$, requiring logarithmic differentiation:
$$\log y = \log [(\log x)^{\cos x}]$$
Using the property $\log(a^b) = b \log a$:
$$\log y = \cos x \cdot \log(\log x)$$
Differentiate both sides w.r.t. $x$ (using the Product Rule on the RHS):
$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x) \cdot \log(\log x) + \cos x \cdot \frac{d}{dx}(\log(\log x))$$
$$\frac{1}{y} \frac{dy}{dx} = (-\sin x) \log(\log x) + \cos x \cdot \left( \frac{1}{\log x} \cdot \frac{1}{x} \right)$$
$$\frac{dy}{dx} = y \left[ -\sin x \log(\log x) + \frac{\cos x}{x \log x} \right]$$
$$\mathbf{\frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} – \sin x \log(\log x) \right]}$$
4. $y = x^x – 2^{\sin x}$
This is the difference of two functions, $u = x^x$ and $v = 2^{\sin x}$.
$$\frac{dy}{dx} = \frac{du}{dx} – \frac{dv}{dx}$$
Find $\frac{du}{dx}$ for $u = x^x$:
$$\log u = \log (x^x) = x \log x$$
$$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$$
$$\frac{du}{dx} = x^x (1 + \log x)$$
Find $\frac{dv}{dx}$ for $v = 2^{\sin x}$:
$$\frac{dv}{dx} = 2^{\sin x} \cdot \log 2 \cdot \frac{d}{dx}(\sin x)$$
$$\frac{dv}{dx} = 2^{\sin x} \cdot \log 2 \cdot \cos x$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = x^x (1 + \log x) – 2^{\sin x} \cos x \log 2}$$
5. $y = (x + 3)^2 (x + 4)^3 (x + 5)^4$
Use logarithmic differentiation:
$$\log y = \log [(x + 3)^2 (x + 4)^3 (x + 5)^4]$$
$$\log y = 2 \log(x+3) + 3 \log(x+4) + 4 \log(x+5)$$
Differentiate both sides w.r.t. $x$:
$$\frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x+3} + 3 \cdot \frac{1}{x+4} + 4 \cdot \frac{1}{x+5}$$
$$\frac{dy}{dx} = y \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right]$$
$$\mathbf{\frac{dy}{dx} = (x + 3)^2 (x + 4)^3 (x + 5)^4 \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right]}$$
6. $y = \left(x + \frac{1}{x}\right)^x + x^{\left(1 + \frac{1}{x}\right)}$
This is the sum of two functions, $u = \left(x + \frac{1}{x}\right)^x$ and $v = x^{\left(1 + \frac{1}{x}\right)}$.
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$
Find $\frac{du}{dx}$ for $u = \left(x + \frac{1}{x}\right)^x$:
$$\log u = x \log \left(x + \frac{1}{x}\right)$$
$$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log \left(x + \frac{1}{x}\right) + x \cdot \frac{1}{x + \frac{1}{x}} \cdot \frac{d}{dx}\left(x + x^{-1}\right)$$
$$\frac{1}{u} \frac{du}{dx} = \log \left(x + \frac{1}{x}\right) + x \cdot \frac{x}{x^2 + 1} \cdot \left(1 – \frac{1}{x^2}\right)$$
$$\frac{1}{u} \frac{du}{dx} = \log \left(x + \frac{1}{x}\right) + \frac{x^2}{x^2 + 1} \cdot \frac{x^2 – 1}{x^2}$$
$$\frac{du}{dx} = \left(x + \frac{1}{x}\right)^x \left[ \log \left(x + \frac{1}{x}\right) + \frac{x^2 – 1}{x^2 + 1} \right]$$
Find $\frac{dv}{dx}$ for $v = x^{\left(1 + \frac{1}{x}\right)}$:
$$\log v = \left(1 + \frac{1}{x}\right) \log x = \log x + \frac{\log x}{x}$$
$$\frac{1}{v} \frac{dv}{dx} = \frac{1}{x} + \frac{x \cdot \frac{1}{x} – \log x \cdot 1}{x^2}$$
$$\frac{1}{v} \frac{dv}{dx} = \frac{1}{x} + \frac{1 – \log x}{x^2} = \frac{x + 1 – \log x}{x^2}$$
$$\frac{dv}{dx} = x^{\left(1 + \frac{1}{x}\right)} \left[ \frac{x + 1 – \log x}{x^2} \right]$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = \left(x + \frac{1}{x}\right)^x \left[ \log \left(x + \frac{1}{x}\right) + \frac{x^2 – 1}{x^2 + 1} \right] + x^{\left(1 + \frac{1}{x}\right)} \left[ \frac{x + 1 – \log x}{x^2} \right]}$$
7. $y = (\log x)^x + x^{\log x}$
This is the sum of two functions, $u = (\log x)^x$ and $v = x^{\log x}$.
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$
Find $\frac{du}{dx}$ for $u = (\log x)^x$:
$$\log u = x \log (\log x)$$
$$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log (\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log (\log x) + \frac{1}{\log x}$$
$$\frac{du}{dx} = (\log x)^x \left[ \log (\log x) + \frac{1}{\log x} \right]$$
Find $\frac{dv}{dx}$ for $v = x^{\log x}$:
$$\log v = (\log x) \log x = (\log x)^2$$
$$\frac{1}{v} \frac{dv}{dx} = 2 (\log x) \cdot \frac{1}{x}$$
$$\frac{dv}{dx} = x^{\log x} \left[ \frac{2 \log x}{x} \right]$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = (\log x)^x \left[ \log (\log x) + \frac{1}{\log x} \right] + x^{\log x} \left[ \frac{2 \log x}{x} \right]}$$
8. $y = (\sin x)^x + \sin^{-1} \sqrt{x}$
This is the sum of $u = (\sin x)^x$ and $v = \sin^{-1} \sqrt{x}$.
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$
Find $\frac{du}{dx}$ for $u = (\sin x)^x$:
$$\log u = x \log (\sin x)$$
$$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \log (\sin x) + x \cot x$$
$$\frac{du}{dx} = (\sin x)^x [\log (\sin x) + x \cot x]$$
Find $\frac{dv}{dx}$ for $v = \sin^{-1} \sqrt{x}$: (Chain Rule)
$$\frac{dv}{dx} = \frac{1}{\sqrt{1 – (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$$
$$\frac{dv}{dx} = \frac{1}{\sqrt{1 – x}} \cdot \frac{1}{2\sqrt{x}}$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = (\sin x)^x [\log (\sin x) + x \cot x] + \frac{1}{2\sqrt{x}\sqrt{1 – x}}}$$
9. $y = x^{\sin x} + (\sin x)^{\cos x}$
This is the sum of $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$.
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$
Find $\frac{du}{dx}$ for $u = x^{\sin x}$:
$$\log u = \sin x \log x$$
$$\frac{1}{u} \frac{du}{dx} = \cos x \cdot \log x + \sin x \cdot \frac{1}{x}$$
$$\frac{du}{dx} = x^{\sin x} \left[ \cos x \log x + \frac{\sin x}{x} \right]$$
Find $\frac{dv}{dx}$ for $v = (\sin x)^{\cos x}$:
$$\log v = \cos x \log (\sin x)$$
$$\frac{1}{v} \frac{dv}{dx} = (-\sin x) \log (\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x$$
$$\frac{dv}{dx} = (\sin x)^{\cos x} \left[ -\sin x \log (\sin x) + \frac{\cos^2 x}{\sin x} \right]$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = x^{\sin x} \left[ \cos x \log x + \frac{\sin x}{x} \right] + (\sin x)^{\cos x} \left[ \frac{\cos^2 x}{\sin x} – \sin x \log (\sin x) \right]}$$
10. $y = x^{\cos x} + \frac{x^2 – 1}{x^2 + 1}$
This is the sum of $u = x^{\cos x}$ and $v = \frac{x^2 – 1}{x^2 + 1}$.
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$
Find $\frac{du}{dx}$ for $u = x^{\cos x}$:
$$\log u = \cos x \log x$$
$$\frac{1}{u} \frac{du}{dx} = (-\sin x) \log x + \cos x \cdot \frac{1}{x}$$
$$\frac{du}{dx} = x^{\cos x} \left[ \frac{\cos x}{x} – \sin x \log x \right]$$
Find $\frac{dv}{dx}$ for $v = \frac{x^2 – 1}{x^2 + 1}$: (Quotient Rule)
$$\frac{dv}{dx} = \frac{(x^2 + 1) \cdot 2x – (x^2 – 1) \cdot 2x}{(x^2 + 1)^2}$$
$$\frac{dv}{dx} = \frac{2x^3 + 2x – (2x^3 – 2x)}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = x^{\cos x} \left[ \frac{\cos x}{x} – \sin x \log x \right] + \frac{4x}{(x^2 + 1)^2}}$$
11. $y = (x \cos x)^x + (x \sin x)^{1/x}$
This is the sum of $u = (x \cos x)^x$ and $v = (x \sin x)^{1/x}$.
$$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$
Find $\frac{du}{dx}$ for $u = (x \cos x)^x$:
$$\log u = x \log (x \cos x) = x (\log x + \log (\cos x))$$
$$\frac{1}{u} \frac{du}{dx} = 1 \cdot (\log x + \log (\cos x)) + x \cdot \left(\frac{1}{x} + \frac{1}{\cos x} (-\sin x)\right)$$
$$\frac{1}{u} \frac{du}{dx} = \log (x \cos x) + x \left(\frac{1}{x} – \tan x\right)$$
$$\frac{du}{dx} = (x \cos x)^x [\log (x \cos x) + 1 – x \tan x]$$
Find $\frac{dv}{dx}$ for $v = (x \sin x)^{1/x}$:
$$\log v = \frac{1}{x} \log (x \sin x) = \frac{1}{x} (\log x + \log (\sin x))$$
$$\frac{1}{v} \frac{dv}{dx} = \left(-\frac{1}{x^2}\right) \log (x \sin x) + \frac{1}{x} \cdot \left(\frac{1}{x} + \frac{1}{\sin x} \cos x\right)$$
$$\frac{1}{v} \frac{dv}{dx} = -\frac{\log (x \sin x)}{x^2} + \frac{1}{x} \left(\frac{1}{x} + \cot x\right)$$
$$\frac{dv}{dx} = (x \sin x)^{1/x} \left[ \frac{1}{x^2} + \frac{\cot x}{x} – \frac{\log (x \sin x)}{x^2} \right]$$
Combine the results:
$$\mathbf{\frac{dy}{dx} = (x \cos x)^x [\log (x \cos x) + 1 – x \tan x] + (x \sin x)^{1/x} \left[ \frac{1 + x \cot x – \log (x \sin x)}{x^2} \right]}$$
🔁 Implicit and Logarithmic Differentiation (Exercises 12 – 15)
12. $x^y + y^x = 1$
This is the sum of $u = x^y$ and $v = y^x$.
$$u + v = 1 \implies \frac{du}{dx} + \frac{dv}{dx} = 0$$
Find $\frac{du}{dx}$ for $u = x^y$:
$$\log u = y \log x$$
$$\frac{1}{u} \frac{du}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x}$$
$$\frac{du}{dx} = x^y \left[ \log x \frac{dy}{dx} + \frac{y}{x} \right]$$
Find $\frac{dv}{dx}$ for $v = y^x$:
$$\log v = x \log y$$
$$\frac{1}{v} \frac{dv}{dx} = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx}$$
$$\frac{dv}{dx} = y^x \left[ \log y + \frac{x}{y} \frac{dy}{dx} \right]$$
Combine and Solve for $\frac{dy}{dx}$:
$$x^y \left[ \log x \frac{dy}{dx} + \frac{y}{x} \right] + y^x \left[ \log y + \frac{x}{y} \frac{dy}{dx} \right] = 0$$
Group $\frac{dy}{dx}$ terms:
$$\frac{dy}{dx} \left[ x^y \log x + y^x \frac{x}{y} \right] = – \left[ x^y \frac{y}{x} + y^x \log y \right]$$
$$\mathbf{\frac{dy}{dx} = – \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}}$$
(Note: $\frac{x}{y} y^x = x y^{x-1}$ and $\frac{y}{x} x^y = y x^{y-1}$)
13. $y^x = x^y$
Use logarithmic differentiation on both sides:
$$\log (y^x) = \log (x^y)$$
$$x \log y = y \log x$$
Differentiate both sides implicitly w.r.t. $x$ (using the Product Rule on both sides):
$$1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x}$$
Group $\frac{dy}{dx}$ terms:
$$\frac{dy}{dx} \left[ \frac{x}{y} – \log x \right] = \frac{y}{x} – \log y$$
$$\frac{dy}{dx} \left[ \frac{x – y \log x}{y} \right] = \frac{y – x \log y}{x}$$
$$\mathbf{\frac{dy}{dx} = \frac{y}{x} \left( \frac{y – x \log y}{x – y \log x} \right)}$$
14. $(\cos x)^y = (\cos y)^x$
Use logarithmic differentiation on both sides:
$$\log ((\cos x)^y) = \log ((\cos y)^x)$$
$$y \log (\cos x) = x \log (\cos y)$$
Differentiate implicitly w.r.t. $x$:
$$\frac{dy}{dx} \cdot \log (\cos x) + y \cdot \frac{1}{\cos x} (-\sin x) = 1 \cdot \log (\cos y) + x \cdot \frac{1}{\cos y} (-\sin y) \frac{dy}{dx}$$
$$\frac{dy}{dx} \log (\cos x) – y \tan x = \log (\cos y) – x \tan y \frac{dy}{dx}$$
Group $\frac{dy}{dx}$ terms:
$$\frac{dy}{dx} [\log (\cos x) + x \tan y] = \log (\cos y) + y \tan x$$
$$\mathbf{\frac{dy}{dx} = \frac{\log (\cos y) + y \tan x}{\log (\cos x) + x \tan y}}$$
15. $xy = e^{(x – y)}$
Use logarithmic differentiation on both sides:
$$\log (xy) = \log (e^{x – y})$$
$$\log x + \log y = x – y$$
Differentiate implicitly w.r.t. $x$:
$$\frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 – \frac{dy}{dx}$$
Group $\frac{dy}{dx}$ terms:
$$\frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 – \frac{1}{x}$$
$$\frac{dy}{dx} \left( \frac{1}{y} + 1 \right) = \frac{x – 1}{x}$$
$$\frac{dy}{dx} \left( \frac{1 + y}{y} \right) = \frac{x – 1}{x}$$
$$\mathbf{\frac{dy}{dx} = \frac{y(x – 1)}{x(1 + y)}}$$
✨ Additional Problems (Exercises 16 – 18)
16. Find the derivative of $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find $f'(1)$.
Use logarithmic differentiation:
$$\log f(x) = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \log(1+x^8)$$
Differentiate w.r.t. $x$:
$$\frac{f'(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$$
$$f'(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right]$$
To find $f'(1)$, first find $f(1)$:
$$f(1) = (1+1)(1+1^2)(1+1^4)(1+1^8) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$$
Now substitute $x=1$ into the derivative expression:
$$\frac{f'(1)}{f(1)} = \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8}$$
$$\frac{f'(1)}{16} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = 0.5 + 1 + 2 + 4 = 7.5$$
$$f'(1) = 16 \times 7.5 = \mathbf{120}$$
17. Differentiate $y = (x^2 – 5x + 8) (x^3 + 7x + 9)$ in three ways.
Let $u = x^2 – 5x + 8$ and $v = x^3 + 7x + 9$.
(i) By using Product Rule
$$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$$
$$\frac{dy}{dx} = (x^2 – 5x + 8) (3x^2 + 7) + (x^3 + 7x + 9) (2x – 5)$$
$$\frac{dy}{dx} = (3x^4 + 7x^2 – 15x^3 – 35x + 24x^2 + 56) + (2x^4 – 5x^3 + 14x^2 – 35x + 18x – 45)$$
$$\frac{dy}{dx} = (3x^4 – 15x^3 + 31x^2 – 35x + 56) + (2x^4 – 5x^3 + 14x^2 – 17x – 45)$$
$$\mathbf{\frac{dy}{dx} = 5x^4 – 20x^3 + 45x^2 – 52x + 11}$$
(ii) By expanding the product to obtain a single polynomial
$$y = x^2(x^3 + 7x + 9) – 5x(x^3 + 7x + 9) + 8(x^3 + 7x + 9)$$
$$y = (x^5 + 7x^3 + 9x^2) – (5x^4 + 35x^2 + 45x) + (8x^3 + 56x + 72)$$
$$y = x^5 – 5x^4 + (7x^3 + 8x^3) + (9x^2 – 35x^2) + (-45x + 56x) + 72$$
$$y = x^5 – 5x^4 + 15x^3 – 26x^2 + 11x + 72$$
Differentiate w.r.t. $x$:
$$\mathbf{\frac{dy}{dx} = 5x^4 – 20x^3 + 45x^2 – 52x + 11}$$
(iii) By logarithmic differentiation
$$\log y = \log(x^2 – 5x + 8) + \log(x^3 + 7x + 9)$$
Differentiate w.r.t. $x$:
$$\frac{1}{y} \frac{dy}{dx} = \frac{2x – 5}{x^2 – 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9}$$
$$\frac{dy}{dx} = (x^2 – 5x + 8) (x^3 + 7x + 9) \left[ \frac{2x – 5}{x^2 – 5x + 8} + \frac{3x^2 + 7}{x^3 + 7x + 9} \right]$$
$$\frac{dy}{dx} = (x^3 + 7x + 9) (2x – 5) + (x^2 – 5x + 8) (3x^2 + 7)$$
This expression is identical to the one obtained in part (i) before expansion.
Therefore, $\mathbf{\frac{dy}{dx} = 5x^4 – 20x^3 + 45x^2 – 52x + 11}$.
Conclusion: Yes, all three methods give the same answer.
18. Proof of Product Rule for three functions: $\frac{d}{dx} (u \cdot v \cdot w)$
First method: Repeated application of the Product Rule
Treat $u \cdot (v \cdot w)$ as a product of two functions, $u$ and $(v \cdot w)$:
$$\frac{d}{dx} (u \cdot (v \cdot w)) = \frac{du}{dx} (v \cdot w) + u \cdot \frac{d}{dx} (v \cdot w)$$
Now, apply the Product Rule to the term $\frac{d}{dx} (v \cdot w)$:
$$\frac{d}{dx} (v \cdot w) = \frac{dv}{dx} w + v \frac{dw}{dx}$$
Substitute this back into the first equation:
$$\frac{d}{dx} (u \cdot v \cdot w) = \frac{du}{dx} v w + u \left( \frac{dv}{dx} w + v \frac{dw}{dx} \right)$$
$$\mathbf{\frac{d}{dx} (u \cdot v \cdot w) = \frac{du}{dx} v w + u \frac{dv}{dx} w + u v \frac{dw}{dx}}$$
Second method: Logarithmic Differentiation
Let $y = u \cdot v \cdot w$.
Take the natural logarithm of both sides:
$$\log y = \log (u \cdot v \cdot w)$$
$$\log y = \log u + \log v + \log w$$
Differentiate both sides w.r.t. $x$:
$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}$$
Multiply both sides by $y = u \cdot v \cdot w$:
$$\frac{dy}{dx} = y \left( \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right)$$
$$\frac{dy}{dx} = (u v w) \left( \frac{1}{u} \frac{du}{dx} \right) + (u v w) \left( \frac{1}{v} \frac{dv}{dx} \right) + (u v w) \left( \frac{1}{w} \frac{dw}{dx} \right)$$
$$\mathbf{\frac{d}{dx} (u \cdot v \cdot w) = \frac{du}{dx} v w + u \frac{dv}{dx} w + u v \frac{dw}{dx}}$$
Both methods yield the same result.