Derivative, the tangent line $y = \sqrt{9-4x}$ at point (-4,5) The function is given,
$$y=\sqrt{9-4x}$$
Now I have to find the tangent line at point (-4,5).
This is how i did it previously...
Derivative; The Tangent line $y=x^2 -4x -5 ; (-2,7)$
Now, I think I'm getting mixed or something...
This is what I did so far:
$$$$
$$f(x)=\sqrt{9-4x}\\f(x_1)=\sqrt{9-4x_1}\\f(x_1+\Delta x) = \sqrt{9-4(x_1+\Delta x})$$
$$$$
So,
$$\lim_{\Delta x\to 0} {f(x_1+\Delta x) - f(x_1) \over \Delta x}
\\=\lim_{\Delta x\to 0} {\sqrt{9-4(x_1 +\Delta x)} - \sqrt{9-4x_1}\over \Delta x}
\\=\lim_{\Delta x\to 0}{\sqrt{9-4x_1-4\Delta x -9 +4x_1}\over \Delta x}
\\=\lim_{\Delta x\to 0}{\sqrt{-4\Delta x} \over \Delta x}$$
Now, I tried doing,
$$\lim_{\Delta x \to 0} {\sqrt{-4\Delta x}\over\sqrt {\Delta x^2}}$$
I'm not sure if it's right, but I can't get it to work.
Any help??
Thanks
 A: 
$$=\lim_{\Delta x\to 0} {\sqrt{9-4(x_1 +\Delta x)} - \sqrt{9-4x_1}\over \Delta x}
\\ \neq \lim_{\Delta x\to 0}{\sqrt{9-4x_1-4\Delta x -9 +4x_1}\over \Delta x}$$

You made the mistake of taking the difference of two roots and making it the square root of the difference. But $$\sqrt a - \sqrt b \neq \sqrt{a-b}$$
Let's go back to the following line: $$=\lim_{\Delta x\to 0} {\sqrt{9-4(x_1 +\Delta x)} - \sqrt{9-4x_1}\over \Delta x}$$ Multiply the numerator and denominator by the conjugate of the numerator, which gives you a difference of squares in the numerator.
$$\begin{align} \lim_{\Delta x\to 0} {f(x_1+\Delta x) - f(x_1) \over \Delta x}
& =\lim_{\Delta x\to 0} {\sqrt{9-4(x_1 +\Delta x)} - \sqrt{9-4x_1}\over \Delta x}\\ \\
&=\lim_{\Delta x\to 0}\;\; \frac{(9-4x_1 - 4\Delta x) - (9-4x_1)}{\Delta x (\sqrt{9-4x_1+4 \Delta x} + \sqrt{9-4x_1})}\\ \\
&=\lim_{\Delta x\to 0}\frac {-4}{\sqrt{9-4x_1 + 4\Delta x} + \sqrt{9-4x_1}}\\ \\
&=  \frac {-4}{2\cdot\sqrt{9-4x_1}}\\ \\
&= \frac {-2}{\sqrt{9-4x_1}}\end{align} $$
Now evaluate the result at $x_1 = -4$. Doing so will give you the slope $m$ of the line tangent at $(-4, 5)$. Using point slope form, that gives us the equation of the line $$y - 5 = m(x +4)$$
A: hint: $$f(x)=\sqrt{9-4x}$$ then we can write $$f(x)=(9-4x)^{1/2}$$ and the first derivative by the power and chaine rule is given by $$f'(x)=\frac{1}{2}(9-4x)^{-1/2}(-4)$$ thus we get $$f'(x)=\frac{-2}{\sqrt{9-4(-4)}}=-\frac{2}{5}$$ and the tangent line is given by $$y=-\frac{2}{5}x+n$$ and for $x=-4$ we get $n=\frac{17}{4}$
for the first derivative with the definition see here
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{9-4(x+h)}-\sqrt{9-x}}{h}=\frac{-4}{\sqrt{9-4(x+h)}+\sqrt{9-4x}}$$ and the limit for $h->0$ is given by $\frac{-2}{\sqrt{9-4x}}$
A: Differentiating the equation wrt. $x$  gives the slope at any $x$ on the curve, and if $y=\sqrt{9-4x}$,
$\dfrac{dy}{dx}=\dfrac{d\sqrt{9-4x}}{dx}=\dfrac{1}{2\sqrt{9-4x}}\cdot\dfrac{d(9-4x)}{dx}$
$\implies \dfrac{-4}{2\sqrt{9-4x}}=\dfrac{-2}{\sqrt{9-4x}}$
Putting, $x=-4$, The slope comes out to be, $\dfrac{-2}{5}.$
Using, $(y-y_1)=m(x-x_1)$, The equation of the line comes out to be,
$(y-5)=\dfrac{-2}{5}(x-(-4))$
$\implies 2x+5y-17=0$
