Question

choose the correct option.​

Answer 1

Answer:

(c) is the answer

Step-by-step explanation:

hope it is helpful for you if it's correct pls mark me in bran list

Answer 2

Solution:

Given that:

$\rm \longrightarrow y = \ln(3x - 2)$

Using chain rule of differentiation, differentiating both sides with respect to x, we get:

$\rm \longrightarrow \dfrac{dy}{dx}= \dfrac{du}{dx} \times \dfrac{d}{du} \ln(u) \: \: \: \big( u =3x - 2 \big)$

$\rm \longrightarrow \dfrac{dy}{dx}= \dfrac{d}{dx}(3x - 2) \times \dfrac{1}{3x - 2}$

$\rm \longrightarrow \dfrac{dy}{dx}= 3 \times \dfrac{1}{3x - 2}$

$\rm \longrightarrow \dfrac{dy}{dx}= \dfrac{3}{3x - 2}$

★ Therefore, option a is the right answer for the problem.

Answer:

$\rm \hookrightarrow \dfrac{dy}{dx}= \dfrac{3}{3x - 2}$

Learn More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$