Question

by applying the chain rule , find the derivative x-1/√x+1=?

Answer 1

Answer:

$\frac{dy}{dx}=\frac{3\sqrt{x} }{2}+\frac{1}{2\sqrt{x} }+1$

Formula :

• $\frac{d}{dx}(uv)=u \frac{dv}{dx}+v\frac{du}{dx}$

• $\frac{d}{dx}(\sqrt{x} ) =\frac{1}{2\sqrt{x} }$

• $\frac{d}{dx} (constant)=0$

To Find :   $\frac{dy}{dx}$   of   $\frac{x+1}{\sqrt{x}+1 }$

Step-by-step explanation:

Let ,

$y=\frac{x+1}{\sqrt{x}+1 } \\\\=>\frac{dy}{dx}= (x+1).\frac{d}{dx}( \sqrt{x}+1)+(\sqrt{x}+1).\frac{d}{dx}(x+1) \\\\=>\frac{dy}{dx}=(x+1)(\frac{1}{2\sqrt{x} } )+(\sqrt{x} +1)(1)\\\\=>\frac{dy}{dx}=\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x} } +\sqrt{x} +1\\\\=>\frac{dy}{dx}=\frac{3\sqrt{x} }{2}+\frac{1}{2\sqrt{x} }+1$

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