Question

Find derivative using first principle definition f(x)=(square root of 9-x)+ ( -2x+1/(x+2))

Answer 1

Explanation:

Here we have the following function:

$f(x)=\sqrt{9-x}+\frac{-2x+1}{x+2}$

By property:

$\left(f\pm g\right)'=f\:'\pm g'$

Then:

$\frac{df}{dx}=\frac{d}{dx}\left(\sqrt{9-x}\right)+\frac{d}{dx}\left(\frac{1-2x}{x+2}\right)$

From here:

$\frac{d}{dx}\left(\sqrt{9-x}\right)=-\frac{1}{2\sqrt{9-x}}$

And:

$\frac{d}{dx}\left(\frac{1-2x}{x+2}\right)=\frac{\frac{d}{dx}[\left(1-2x\right)]\left(x+2\right)-\frac{d}{dx}[\left(x+2\right)]\left(1-2x\right)}{\left(x+2\right)^2} \\ \\ =\frac{\left(-2\right)\left(x+2\right)-1\cdot \left(1-2x\right)}{\left(x+2\right)^2}= -\frac{5}{\left(x+2\right)^2}$

Therefore:

$\boxed{\frac{df}{dx}=-\frac{1}{2\sqrt{9-x}}-\frac{5}{\left(x+2\right)^2}}$