Question

if y=log(√x+√x-a) then dy/dx is​

Answer 1

$\boxed{\frac{dy}{dx}=\frac{1}{2\ln 10\sqrt{x(x-a})}}$

Step-by-step explanation:

Given:

$y=\log(\sqrt{x}+\sqrt{x-a})$

To find out:

Derivative of y

Solution:

Let $\sqrt{x}+\sqrt{x-a}=t$

$\frac{dt}{dx}=\frac{d}{dx}(\sqrt{x}+\sqrt{x-a})$

$\implies \frac{dt}{dx}=\frac{d}{dx}(\sqrt{x})+\frac{d}{dx}(\sqrt{x-a})$

$\implies \frac{dt}{dx}=\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x-a}}$

$\implies \frac{dt}{dx}=\frac{\sqrt{x}+\sqrt{x-a}}{2\sqrt{x}\sqrt{x-a}}$

Now

$y=\log t$

$\frac{dy}{dx}=\frac{d}{dx}(\log t)$

$\implies \frac{dy}{dx}=\frac{d}{dt}(\log t).\frac{dt}{dx}$

$\implies \frac{dy}{dx}=\frac{1}{t\ln 10}.(\frac{\sqrt{x}+\sqrt{x-a}}{2\sqrt{x}\sqrt{x-a}})$

$\implies \frac{dy}{dx}=\frac{1}{(\sqrt{x}+\sqrt{x-a})\ln 10}.(\frac{\sqrt{x}+\sqrt{x-a}}{2\sqrt{x}\sqrt{x-a}})$

$\implies \frac{dy}{dx}=\frac{1}{2\ln 10\sqrt{x(x-a})}$

Hope this answer is helpful.

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