Can someone please expain to me how to find the derivative of this function: (square root of (3-2x)) Thank you!
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Originally Posted by BubblegumPops Can someone please expain to me how to find the derivative of this function: (square root of (3-2x)) Thank you! $\displaystyle \sqrt{3-2x} = \sqrt{t}$ where $\displaystyle t=3-2x$. By chain rule, $\displaystyle \frac{dy}{dt} = \frac{1}{2}t^{-1/2} , \frac{dx}{dt} = -2 \mbox{ so }\frac{dy}{dx} = \frac{1}{2} (-2) t^{-1/2} = - (3-2x)^{-1/2}$
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