Question

for which of the following functions is the chain rule an appropriate method to find the derivative with respect to x?

1. ( y=sin left(3 x^{2}

ight) )
ii. ( y=e^{x} \tan x )
iii. ( y=\frac{1}{8 x^{4}-2 x} )

Explanation:

Step1: Analyze function I

For \(y = \sin(3x^{2})\), let \(u = 3x^{2}\), then \(y=\sin(u)\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). \(\frac{dy}{du}=\cos(u)\) and \(\frac{du}{dx}=6x\), so \(\frac{dy}{dx}=6x\cos(3x^{2})\).

Step2: Analyze function II

For \(y = e^{x}\tan x\), by the product rule \((uv)^\prime=u^\prime v + uv^\prime\) where \(u = e^{x}\), \(u^\prime=e^{x}\), \(v=\tan x\), \(v^\prime=\sec^{2}x\). But if we consider the composition in a wrong way (not really, but for the sake of chain - rule check), it is a product of two functions, not a composition in the sense of \(y = f(g(x))\). However, if we rewrite \(y\) as \(y = e^{x}\cdot\tan x\), it's a product. But if we consider \(\tan x=\frac{\sin x}{\cos x}\), and \(e^{x}\) is separate. But actually, the chain - rule is for composite functions \(y = f(g(x))\). Here, it's a product. But wait, if we consider \(y = e^{x}\tan x\) as \(y = e^{x}\cdot\tan x\), we use the product rule \((uv)^\prime = u^\prime v+uv^\prime\) (\(u = e^{x}\), \(u^\prime=e^{x}\), \(v = \tan x\), \(v^\prime=\sec^{2}x\)). But if we force - fit: Let \(u = e^{x}\) and \(v=\tan x\), it's not a composite function. However, if we consider \(y\) as \(y=e^{x}\cdot\tan x\), the chain - rule is not the primary rule (product rule is). But wait, actually, no, the chain - rule is for \(y = f(g(x))\). For \(y = e^{x}\tan x\), it's a product. But wait, no, if we write \(y\) as \(y = e^{x}\cdot\tan x\), we use the product rule \((uv)^\prime=u^\prime v + uv^\prime\) (\(u = e^{x}\), \(u^\prime=e^{x}\), \(v=\tan x\), \(v^\prime=\sec^{2}x\)). But if we consider \(y\) as a composition? No. Wait, actually, for \(y=\sin(3x^{2})\), it is \(y = f(g(x))\) where \(f(u)=\sin(u)\) and \(g(x)=3x^{2}\). For \(y=\frac{1}{8x^{4}-2x}\), we can write \(y=(8x^{4}-2x)^{- 1}\), let \(u = 8x^{4}-2x\), then \(y = u^{-1}\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), \(\frac{dy}{du}=-u^{-2}\) and \(\frac{du}{dx}=32x^{3}-2\), so \(\frac{dy}{dx}=\frac{-(32x^{3}-2)}{(8x^{4}-2x)^{2}}\)

Answer:

I and III.