if y=(√3x^2-4), then dy = a. 6x(√3x^2-4)dx b. (3x)/(√3x^2-4)dx c. (6x)/(√3x^2-4) d.
2(√3x^2-4)/(3x)dx

there is an E choice but it cut off, is it E??...

Question

if y=(√3x^2-4), then dy = a. 6x(√3x^2-4)dx  b.  (3x)/(√3x^2-4)dx c. (6x)/(√3x^2-4)    d. 
2(√3x^2-4)/(3x)dx

there is an E choice but it cut off, is it E??...

anonymous · Answer

Dude, we're not here to do problems for you.  I'm here to help work through them.  Your question should be:  Help me find the differential dy given
$$y=\sqrt x^2-4$$

anonymous · Answer

So do you need help differentiating or not?  Also, your function's not clear.  Is this it?
$$y=\sqrt{3x^2-4}$$

anonymous · Answer

yes vf321, sorry but $$y=\sqrt{3x^2-4}$$ is right

anonymous · Answer

When dealing with chain rule, it often helps to make your own functions up. Let:
$$f(x) = \sqrt x$$$$g(x)=3x^2-4$$Then,$$y=f(g(x))$$Well let's take d/dx of both sides.$$\frac{dy}{dx}=f'(g(x))g'(x)$$Now, all you have to do is multiply by dx on both sides and replace all the f's and g's with their actual values.
$$\frac{dy}{dx}dx=f'(g(x))g'(x)dx$$$$dy = f'(g(x))g'(x)dx$$