Question

A level differentiation question. Mainly stuck on the 4-x/x part. Help would be appreciated.

Answer 1

Answer:

see explanation

Step-by-step explanation:

Differentiate $\frac{4-x}{x}$ using the quotient rule, given

y = $\frac{f(x)}{g(x)}$ , then

$\frac{dy}{dx}$ = $\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

Here f(x) = 4 - x ⇒ f'(x) = - 1

g(x) = x ⇒ g'(x) = 1 , thus

$\frac{dy}{dx}$ = $\frac{-x-(4-x)}{x^2}$ = $\frac{-x-4+x}{x^2}$ = - $\frac{4}{x^2}$

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Given

y = 3x² + $\frac{4-x}{x}$, then

$\frac{dy}{dx}$ = 6x - $\frac{4}{x^2}$ ← evaluate for x = 2

$\frac{dy}{dx}$ = 6(2) - $\frac{4}{4}$ = 12 - 1 = 11 ← as required

Let me know if you require assistance on (b) and (c)