Question

y=18 x+log(x-4) find dy/dx​

Answer 1

Answer

$\sf \dashrightarrow \dfrac{18x-71}{x-4}$

Given :

• y = 18x + log(x - 4)

Identity to be used :

$\sf If \: z = f(y) \: and \: \: y = g(x) \: then \\ \sf\longrightarrow \dfrac{dz}{dy} = \dfrac{dz}{dy}.\dfrac{dy}{dx} \\\\ \sf \longrightarrow \dfrac{d}{dx}(\log x)=\dfrac{1}{x} \\\\ \sf \longrightarrow \dfrac{d}{dx}(constant) = 0$

To Find :

• The derivative of y

Solution :

$\sf y = 18x + \log(x - 4)\\\\ \sf \implies \dfrac{dy}{dx}=\dfrac{d}{dx} \{18x + log(x - 4) \} \\\\ \sf \implies \dfrac{dy}{dx}=\dfrac{d}{dx}(18x) + \dfrac{d}{dx} \{\log (x - 4) \} \\\\ \sf \implies \dfrac{dy}{dx} = 18 + \{ \dfrac{1}{x-4} \}.\{ \dfrac{d}{dx} (x - 4) \} \\\\ \sf \implies \dfrac{dy}{dx} = 18 + \dfrac{1}{x-4} \\\\ \sf \implies \dfrac{dy}{dx} = \dfrac{18(x -4)+ 1}{x - 4} \\\\ \sf \implies \dfrac{dy}{dx} = \dfrac{18x - 71}{x - 4}$