Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

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?

share|improve this answer
    
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
add comment

Let $f(x)=\sqrt{g(x)}$ , then using chain rule we have :

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.