Question

differentiate pls solve​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=\dfrac{\sqrt{x^2-a^2}}{x}}$

$\sf{Put\,\,\,x=a\,sec(\theta)}$

Differentiating it w.r.t. θ

$\sf{\implies\,\dfrac{dx}{d\theta}=a\,sec(\theta)\,tan(\theta)\,\,\,\,\,\,\,\,...(1)}$

Now,

$\sf{y=\dfrac{\sqrt{(a\,sec(\theta))^2-a^2}}{a\,sec(\theta)}}$

$\sf{\implies\,y=\dfrac{\sqrt{a^2\,sec^2(\theta)-a^2}}{a\,sec(\theta)}}$

$\sf{\implies\,y=\dfrac{\sqrt{a^2\,(sec^2(\theta)-1)}}{a\,sec(\theta)}}$

$\sf{\implies\,y=\dfrac{a\sqrt{(sec^2(\theta)-1)}}{a\,sec(\theta)}}$

$\sf{\implies\,y=\dfrac{\sqrt{tan^2(\theta)}}{sec(\theta)}}$

$\sf{\implies\,y=\dfrac{tan(\theta)}{sec(\theta)}}$

$\sf{\implies\,y=sin(\theta)}$

Differentiating it w.r.t. θ

$\sf{\implies\,\dfrac{dy}{d\theta}=cos(\theta)\,\,\,\,\,\,\,\,\,...(2)}$

Divide (2) by (1),

$\sf{\dfrac {dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{cos(\theta)}{a\,sec(\theta)\,tan(\theta)}}$

$\sf{\implies\dfrac {dy}{dx}=\dfrac{1}{a\,sec^2(\theta)\,tan(\theta)}}$

$\sf{Since,\,\,\,sec(\theta)=\dfrac{x}{a}}\\\\\sf{So,\,\,\,tan(\theta)=\sqrt{\bigg(\dfrac{x}{a}\bigg)^2-1}}$

$\sf{\implies\dfrac {dy}{dx}=\dfrac{1}{a\,\bigg(\dfrac{x}{a}\bigg)^2\,\sqrt{\bigg(\dfrac{x}{a}\bigg)^2-1}}}$

$\sf{\implies\dfrac {dy}{dx}=\dfrac{a}{x^2\,\sqrt{\dfrac{x^2-a^2}{a^2}}}}$

$\sf{\implies\dfrac {dy}{dx}=\dfrac{a^2}{x^2\,\sqrt{x^2-a^2}}}$