Question

Differentiate the following equation with respect to x:y=loge (ax+b)
X के सबध में निम्नलिखित समीकरण को शिन करे: \uge (ax b)​

Answer 1

Answer:

$\bold{\frac{dy}{dx}=\frac{a}{ax+b}}$

Step-by-step explanation:

I have applied chain rule to find derivative of the given function.

Given:

$y=log_e(ax+b)$

Take $u=ax+b$

Then $\frac{du}{dx}=a.1+0$

$\implies\:\frac{du}{dx}=a$

Now,

$y=log_eu$

$\frac{dy}{du}=\frac{1}{u}$

By chain rule,

$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}$

$\frac{dy}{dx}=\frac{1}{u}.a$

$\frac{dy}{dx}=\frac{1}{ax+b}.a$

$\implies\:\frac{dy}{dx}=\frac{a}{ax+b}$