rate of change dy/dx

[Becky4paws]

Find rate of change for prescribed value of x.

y=x+(3/2-4x) x=0
y=x +3/2 - 3/4x

I don't think I've even started out correctly. I'm not sure how to find the derivative of 3/4x. I think it would be 3/4 x ^-1/4. Help.

 

[pka]

Please edit your post to make it clear.
Is it (3/2)-(1/4x) or 3/(2-4x)?

 

[Becky4paws]

Sorry-3/(2-4x)

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
y & = & x + \frac{3}{{2 - 4x}} = x + 3\left( {2 - 4x} \right)^{ - 1} \\
y' & = & 1 - 3( - 4)(2 - 4x)^{ - 2} = 1 + \frac{{12}}{{(2 - 4x)^2 }} \\
y'(0) & = & ? \\
\end{array}\)

 

[Becky4paws]

y'(0) = 1

 

[ChaoticLlama]

y'(0) = 1

Try again. Show your steps regarding how you solve for y'(0).

 

[pka]

y'(0) = 1

Try again:

\(\displaystyle \L
y' = 1 + \frac{{12}}{{(2 - 4x)^2 }}\quad \Rightarrow \quad y'(0) = ?\)

 

[Becky4paws]

agghhh...

Ok. I did try to shorten things a bit.

1 + 12/(2-0)^2=

1 +12/4=
1+3=4