Question

Differentiate y = a-x ÷ a + x​

Answer 1

Answer:

Assuming a is a constant the derivative is the constant a. Letting f(x)=ax, f(x)-f(a)=a(x-a). Hence [f(x)-f(a)]/[x-a]=a and taking this limit as x->a equals a and that is the definition of the derivative

Answer 2

ANSWER:

To Differentiate:

• y = (a - x)/(a + x)

Solution:

We are given that,

$\implies y=\dfrac{a-x}{a+x}$

On differentiating,

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{a-x}{a+x}\right)$

Now we will apply Quotient rule, which says,

$\implies y=\dfrac{u}{v}$

$\implies \dfrac{dy}{dx}=\dfrac{vu'-uv'}{v^2}$

(v' and u' means, derivative of v and derivative of u respectively.)

So,

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{a-x}{a+x}\right)$

$\implies \dfrac{dy}{dx}=\dfrac{(a+x)(a-x)'-(a-x)(a+x)'}{(a+x)^2}$

$\implies \dfrac{dy}{dx}=\dfrac{(a+x)(a'-x')-(a-x)(a'+x')}{(a+x)^2}$

We know that,

$\hookrightarrow \dfrac{d}{dx}a=0$

And,

$\hookrightarrow \dfrac{d}{dx}x=1$

So,

$\implies \dfrac{dy}{dx}=\dfrac{(a+x)(a'-x')-(a-x)(a'+x')}{(a+x)^2}$

$\implies \dfrac{dy}{dx}=\dfrac{(a+x)(0-1)-(a-x)(0+1)}{(a+x)^2}$

$\implies \dfrac{dy}{dx}=\dfrac{(a+x)(-1)-(a-x)(1)}{(a+x)^2}$

So,

$\implies \dfrac{dy}{dx}=\dfrac{-a-x-a+x}{(a+x)^2}$

Therefore,

$\implies\bf\dfrac{dy}{dx}=\dfrac{-2a}{(a+x)^2}$