Question

If u, v and w are functions of x, then show that, ${\displaystyle \frac{d}{dx}}(u\cdot v\cdot w)={\displaystyle \frac{du}{dv}}\cdot v\cdot w+u\cdot {\displaystyle \frac{dv}{dx}}\cdot w+u\cdot v\cdot {\displaystyle \frac{dw}{dx}}$ in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Answer 1

Hint: In this question, we need to prove the formula of finding the derivative of the product of three functions using two methods. For repeated application of product rule, we will first consider two terms as one and apply product rule on obtained two terms. After that, we will apply product rule on that term having two terms. Product rule on any two function f(x) and g(x) is given by ${\displaystyle \frac{d}{dx}}f\left(x\right)\cdot g\left(x\right)=f\left(x\right)g^{\prime }\left(x\right)+f^{\prime }\left(x\right)g\left(x\right)$ where f'(x) and g'(x) denotes derivatives of f(x) and g(x) respectively. For logarithmic differentiation, we will consider the product as y and then take log both sides. After that, we will apply $\log (m\cdot n)=\log m+\log n$ to separate three terms and hence find differentiation of logarithmic terms. Derivatives of any logarithmic function is given by, ${\displaystyle \frac{d}{dx}}\log y={\displaystyle \frac{1}{y}}{\displaystyle \frac{dy}{dx}}$.

Complete step-by-step answer:
Let us first find the derivative of the product of three function u, v and w using repeated application of product rule.
Let us suppose function as y. So, we get: $y=u\cdot v\cdot w$.
Taking differentiation both sides w.r.t. x, we get: ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}(u\cdot v\cdot w)$.
Let us consider uv as one term and then apply product rule on uv and w. Product rule on any two functions f(x) and g(x) is given by, ${\displaystyle \frac{d}{dx}}f\left(x\right)\cdot g\left(x\right)=f\left(x\right)\cdot {\displaystyle \frac{d}{dx}}g\left(x\right)+g\left(x\right)\cdot {\displaystyle \frac{d}{dx}}f\left(x\right)$.
Hence we get: ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d\left(uv\right)}{dx}}\cdot w+{\displaystyle \frac{dw}{dx}}\cdot \left(uv\right)$.
Now, we need to find derivative of uv. Using product rule ${\displaystyle \frac{d}{dx}}\left(uv\right)={\displaystyle \frac{udv}{dx}}+{\displaystyle \frac{vdu}{dx}}$ we get:
$$\begin{array}{rl}& {\displaystyle \frac{dy}{dx}}=(u{\displaystyle \frac{dv}{dx}}+v{\displaystyle \frac{du}{dx}})w+{\displaystyle \frac{dw}{dx}}\left(uv\right)\\ & \Rightarrow {\displaystyle \frac{dy}{dx}}=uw{\displaystyle \frac{dv}{dx}}+vw{\displaystyle \frac{du}{dx}}+uv{\displaystyle \frac{dw}{dx}}\end{array}$$
Now y was supposed to be u.v.w, so we get:
$$\Rightarrow {\displaystyle \frac{d}{dx}}(u\cdot v\cdot w)=uw{\displaystyle \frac{dv}{dx}}+vw{\displaystyle \frac{du}{dx}}+uv{\displaystyle \frac{dw}{dx}}$$
Rearranging the terms, we get:
$$\Rightarrow {\displaystyle \frac{d}{dx}}(u\cdot v\cdot w)={\displaystyle \frac{du}{dx}}\cdot v\cdot w+u\cdot {\displaystyle \frac{dv}{dx}}\cdot w+u\cdot v\cdot {\displaystyle \frac{dw}{dx}}$$
Hence proved.
Now, let us prove it using logarithmic differentiation.
We know, $y=u\cdot v\cdot w$.
Taking log on both sides, we get: $\log y=\log \left(uvw\right)$.
We know that $\log \left(ab\right)=\log a+\log b$ hence, we get: $\log y=\log u+\log v+\log w$.
Differentiating both sides w.r.t. x, we get:
$$\begin{array}{rl}& {\displaystyle \frac{d}{dx}}(\log y)={\displaystyle \frac{d}{dx}}(\log u+\log v+\log w)\\ & \Rightarrow {\displaystyle \frac{d}{dx}}(\log y)={\displaystyle \frac{d}{dx}}\log u+{\displaystyle \frac{d}{dx}}\log v+{\displaystyle \frac{d}{dx}}\log w\end{array}$$
As we know that, differentiation of logarithmic function is given by ${\displaystyle \frac{d}{dx}}\log p={\displaystyle \frac{1}{p}}{\displaystyle \frac{dp}{dx}}$.
Hence applying this we get:
$$\Rightarrow {\displaystyle \frac{1}{y}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}+{\displaystyle \frac{1}{v}}{\displaystyle \frac{dv}{dx}}+{\displaystyle \frac{1}{w}}{\displaystyle \frac{dw}{dx}}$$
Taking y on the right side we get:
$$\Rightarrow {\displaystyle \frac{dy}{dx}}=y({\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}+{\displaystyle \frac{1}{v}}{\displaystyle \frac{dv}{dx}}+{\displaystyle \frac{1}{w}}{\displaystyle \frac{dw}{dx}})$$
As we know, y = uvw so putting value, we get:
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{d(u\cdot v\cdot w)}{dx}}=\left(uvw\right)({\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}+{\displaystyle \frac{1}{v}}{\displaystyle \frac{dv}{dx}}+{\displaystyle \frac{1}{w}}{\displaystyle \frac{dw}{dx}})\\ & \Rightarrow {\displaystyle \frac{d(u\cdot v\cdot w)}{dx}}={\displaystyle \frac{uvw}{u}}{\displaystyle \frac{du}{dx}}+{\displaystyle \frac{uvw}{v}}{\displaystyle \frac{dv}{dx}}+{\displaystyle \frac{uvw}{w}}{\displaystyle \frac{dw}{dx}}\\ & \Rightarrow {\displaystyle \frac{d(u\cdot v\cdot w)}{dx}}=vw{\displaystyle \frac{du}{dx}}+uw{\displaystyle \frac{dv}{dx}}+uv{\displaystyle \frac{dw}{dx}}\end{array}$$
Rearranging we get:
$$\Rightarrow {\displaystyle \frac{d}{dx}}(u\cdot v\cdot w)={\displaystyle \frac{du}{dx}}\cdot v\cdot w+u\cdot {\displaystyle \frac{dv}{dx}}\cdot w+u\cdot v\cdot {\displaystyle \frac{dw}{dx}}$$
Hence proved.

Note: Students should take care while applying repeating product rule. They can consider any two terms as one term first and then solve. Take care while applying logarithmic function on both sides and separating terms. Students can forget to put ${\displaystyle \frac{1}{y}}$ on left side of the equation. Always remember that, in the product rule, we use positive sign only. In quotient rule, we use negative sign.