Question

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

Answer 1

$\textbf{\underline{product rule}}$

we know according to product rule,
$\frac{d}{dx}(A.B)=A\frac{dB}{dx}+B\frac{dA}{dx}$ use it here,

$\frac{d}{dx}(u.v.w)=\frac{d}{dx}[(u.v).w]\\\\=(u.v)\frac{dw}{dx}+w\frac{d}{dx}(u.v)\\\\=(u.v)\frac{dw}{dx}+w.[u\frac{dv}{dx}+v\frac{du}{dx}]\\\\=(u.v)\frac{dw}{dx}+(w.u)\frac{dv}{dx}+(w.v)\frac{du}{dx}$

$\textbf{\underline{logrithmn method}}$
taking log on both sides, we get as y = u.v.w
e.g., logy = log(u.v.w)
logy = logu + logv + logw
now differentiate with respect to x
$\frac{1}{y}.\frac{dy}{dx}=\frac{1}{u}.\frac{du}{dx}+\frac{1}{v}.\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\\\=\frac{dy}{dx}=\frac{y}{u}\frac{du}{dx}+\frac{y}{v}.\frac{dv}{dx}+\frac{y}{w}.\frac{dw}{dx}$
now put y = u.v.w
then, $\frac{d}{dx}(u.v.w)=(v.w)\frac{du}{dx}+(w.u)\frac{dv}{dx}+(u.v)\frac{dw}{dx}$

Answer 2

Step-by-step explanation:

hope this answer helps u