# find the derivative respect to $x$ and explain step by step?

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 $\dfrac{-3}{2\sqrt{4-3x}}$

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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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all clear thx! @BrianM.Scott – Sb Sangpi Apr 4 '12 at 6:30
@SbSangpi: You’re very welcome! – Brian M. Scott Apr 4 '12 at 6:32

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! :D – Sb Sangpi Apr 4 '12 at 6:25