Find : $ dy/dx, y=\sqrt{4x^2 - 7x - 2}$ The problem says Find $dy/dx,  y=\sqrt{4x^2 - 7x - 2}$
So far I changed it to $(4x^2 - 7x - 2)^{1/2}$
I don't know where to go from there.
 A: Another option is to use implicit differentiation --- the others seem to assume you know the Chain Rule so no point going down that route (although I use it here).
$$\begin{align}
y(x)&=\sqrt{4x^2-7x-2}
\\ \Rightarrow [y(x)]^2&=4x^2-7x-2.
\end{align}$$
Now I will use the Chain Rule on the LHS while I differentiate both sides with respect to $x$:
$$\begin{align}
2\cdot y(x)\cdot\frac{dy}{dx}&=8x-7
\\ \Rightarrow \frac{dy}{dx}&=\frac{8x-7}{2y(x)}
\\&=\frac{8x-7}{2\sqrt{4x^2-7x-2}}.
\end{align}$$
A: hint: Use the chain rule: $\sqrt{u(x)}' = \dfrac{u'(x)}{2\sqrt{u(x)}}$
A: $$
y = u^{1/2},\qquad u = 4x^2-7x-2,\qquad \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.
$$
A: If $y = \sqrt{4x^2 - 7x - 2}$
then 
$$\frac{dy}{dx} = \frac{d}{dx}(4x^2 - 7x - 2)^{\frac{1}{2}}
\\
$$
$$
=\frac{1}{2}\frac{d}{dx}(4x^2 - 7x - 2) \times (4x^2 - 7x - 2)^{\frac{-1}{2}}
$$
$$
= \frac{8x-7}{2\sqrt{4x^2 - 7x - 2}}
$$
using the chain rule.
That is, if we have a function $y = u^{\frac{1}{2}}$ then $$\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx}$$
