Question

1.To differentiate composite function:-
(i) y = log(ax +b)​

Answer 1

Answer:

  $\frac{d}{dx} log(ax+b)=\frac{a}{ax+b}$

Step-by-step explanation:

Given a function, $y=log(ax+b)$

Differentiating with respect to $x$, we get

                                      $\frac{dy}{dx} =\frac{d}{dx} log(ax+b)$                            

Since, $logx=\frac{1}{x}$

                                      $\frac{dy}{dx} =\frac{1}{ax+b} \frac{d}{dx} (ax+b)$

Since, $\frac{d}{dx} (ax+b)=a$, therefore

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

Answer 2

Given function is

$y = log(ax + b)$

We need to differentiate the given function w.r.t x on both sides.

Differentiation:

Differentiation is a process of finding instantaneous rate of change of a function at a particular point.

If the function is a straight line we can calculate the change of y with respect to the change of x which is same throughout the function.

$slope = delta \: y \div delta \: x$

But if the function is a curve, the change of y with respect to the change of x varies from point to point. In these kind of situations we need to find the instantaneous rate of change at a particular point.

This instantaneous rate of change of y with respect to that of x is known as differentiation. This instantaneous change is very small and this small change is denoted by d.

$differentiation \: = dy \div dx$

Now, we need to find the differentiation of the above function.

$y = log(ax + b)$

Generally we know that

$d \div dx( log(x) ) = 1 \div x$

But in the function, in x place we have

$ax + b$

So, we have to differentiate the log function first. Then apply chain rule to differentiate

$ax + b$

$d \div dx( log(ax + b) ) \\ = 1 \div (ax + b)(d \div dx(ax + b)) \\ = ( 1 \div (ax + b))(a) \\ = a \div (ax + b)$

Note:

Note: we know that differentiation of x is 1 and constant is zero.

Hence,

$d \div dx(ax + b) = a(1) + 0 \\ = a$

Therefore the differentiation of the given function is

$a \div (ax + b)$

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