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Let $y = \sqrt{4-3x}$.

The problem is to find $\dfrac{d}{dx} \sqrt{4-3x}$, i.e. to find $\dfrac{dy}{dx}$.

Let $u=4-3x$. Then $y=\sqrt{u}$.

Then we have $ \frac{dy}{du} = ? or \frac{1}{2}(4-3x)^\frac{-1}{2},\qquad \text{and}\qquad \frac{du}{dx} = -3. $

Therefore $ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = $

the right answer is $\dfrac{-3}{2\sqrt{4-3x}}$

Can you please help me out? thanks

2 Answers 2

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What you have is correct: if you put the pieces together, you get

$\frac{dy}{du}=\frac12u^{-1/2}=\frac12(4-3x)^{-1/2}=\frac1{2\sqrt{4-3x}}\;,$ and $\frac{dy}{dx}=-3\;,$ so by the chain rule $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\left(\frac1{2\sqrt{4-3x}}\right)(-3)=\frac{-3}{2\sqrt{4-3x}}\;.$ If some part of that is not clear, can you explain exactly where the difficulty is?

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    @SbSangpi: You’re very welcome!2012-04-04
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Let $f(x)=\sqrt{g(x)}$ , then using chain rule we have :

f'(x)=\frac{1}{2\sqrt {g(x)}} \cdot g'(x)

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    thx! can u plz solve the quation in that method? I would like to know the steps! :D2012-04-04